Proof. See Proposition 2.19, Lemma 2.31, Corollary C, and Proposition 5.6 in [5]. ◻

Proof. Let \(\varphi:G\to \mathrm{PGL}_q(\mathbb{F})\) be a projective representation, and let \(\alpha_\varphi\) denote its obstruction cocycle, as in Theorem 1. Conjugating \(\varphi\) by an element of \(\mathrm{PGL}_q(\mathbb{F})\) clearly does not change \(\alpha_\varphi\).

The obstruction is also additive under tensor product. Indeed, if \(\psi:G\to \mathrm{PGL}_r(\mathbb{F})\) is another projective representation, then choosing lifts of \(\varphi\) and \(\psi\) gives \[\alpha_{\varphi\otimes\psi}=\alpha_\varphi+\alpha_\psi,\] where the coefficient group \(\mathbb{F}^\times\) is written additively. In particular, tensoring with a linearizable projective representation does not change the obstruction class.

Thus conditions \((1)\)–\((4)\) are all equivalent to the vanishing of the obstruction index. By Theorem 1, this is equivalent to linearizability.

The same obstruction may be defined for a homomorphism \(G\to \mathrm{PGL}_\otimes(\mathbb{F})\), using the central extension \[1\longrightarrow \mathbb{F}^\times \longrightarrow \mathrm{GL}_\otimes(\mathbb{F})\longrightarrow \mathrm{PGL}_\otimes(\mathbb{F})\longrightarrow 1 .\]

Its vanishing is equivalent to liftability to \(\mathrm{GL}_\otimes(\mathbb{F})\), by the same argument as in the finite-dimensional case. Moreover, for a projective representation \(\varphi:G\to \mathrm{PGL}(V)\), the obstruction class is the same as the obstruction class of the induced map \(G\to \mathrm{PGL}_\otimes(\mathbb{F})\). Therefore the stabilized conditions \((5)\) and \((6)\) are again equivalent to the vanishing of the same index, and hence to linearizability. ◻

Proof. Since \(\varphi\) is linearly onsite, it fits in (2 ) with \(\beta = \operatorname{id}\). We can extend the diagram to consider the maps: \[\begin{tikzcd} & {\prod_{x\in X} \operatorname{GL}_{q_x}(\mathbb{F})} & {\operatorname{GL}_{\otimes}(X; \mathbb{F})} \\ & {\prod_{x\in X} \operatorname{PGL}_{q_x}(\mathbb{F})} & {\operatorname{PGL}(X; \mathbb{F})} \\ G & {\mathcal{Q}(X,q)} & {\mathcal{Q}(X; \mathbb{F})} \arrow[from=1-2, to=1-3] \arrow[two heads, from=1-2, to=2-2] \arrow[from=1-3, to=2-3] \arrow[from=2-2, to=2-3] \arrow[hook, from=2-2, to=3-2] \arrow[from=2-3, to=3-3] \arrow[shift left=2, dashed, from=3-1, to=1-2] \arrow[dashed, from=3-1, to=2-2] \arrow["\varphi"', from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \end{tikzcd}.\] The natural map \(\operatorname{BGL}_{\otimes}(X; {\mathbb{F}}) \to \operatorname{BPGL}(X; {\mathbb{F}}) \to \operatorname{BPGL}(X;{\mathbb{F}})^{+}\) sends the commutator subgroup to \(0\) by Lemma 22. The map \(\operatorname{BPGL}(X; {\mathbb{F}}) \to B\mathcal{Q}(X; {\mathbb{F}}) \to B\mathcal{Q}(X; {\mathbb{F}})^+\) also sends the commutator subgroup of \(\operatorname{PGL}(X; {\mathbb{F}})\) to \(0\). Thus, the universal property of plus-construction gives a commutative diagram: \[\begin{tikzcd} BG && \\ {\operatorname{BGL}_{\otimes}(X; \mathbb{F})} & {\operatorname{BPGL}(X; \mathbb{F})} & {B\mathcal{Q}(X; \mathbb{F})} \\ {\operatorname{BGL}_{\otimes}(X; \mathbb{F})^+} & {\operatorname{BPGL}(X; \mathbb{F})^+} & {B\mathcal{Q}(X; \mathbb{F})^+} \arrow[from=1-1, to=2-1] \arrow[from=1-1, to=2-2] \arrow[from=1-1, to=2-3] \arrow[from=2-1, to=2-2] \arrow[from=2-1, to=3-1] \arrow[from=2-2, to=2-3] \arrow[from=2-2, to=3-2] \arrow[from=2-3, to=3-3] \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \end{tikzcd}\] Thus, we see that the stabilization \(BG \to B\mathcal{Q}(X; {\mathbb{F}})^+\) factors through \(\operatorname{BGL}_{\otimes}(X; {\mathbb{F}})^+\). We claim any map \(f: BG \to \operatorname{BGL}_{\otimes}(X; {\mathbb{F}})^+\) is null-homotopic. Indeed, since \(G\) is rationally acyclic, we observe that \(\operatorname{Hom}_{Grp}(G, V) = 0\) for any \(V\) that is rational, as \(V\) is torsion-free. Thus, any map \(BG \to \operatorname{BGL}_{\otimes}(X; {\mathbb{F}})^+\) lifts to the universal cover \(\widetilde{\operatorname{BGL}_{\otimes}(X; {\mathbb{F}})^+}\). We now show that any map \(BG \to \tau_{>1} \operatorname{BGL}_{\otimes}(X; {\mathbb{F}})^+\) is null-homotopic. A null-homotopy is equivalent to being able to lift the map to along the path-space fibration. By standard obstruction theory (e.g., Theorem 7.37 of [32], which we can apply as the universal cover is simply-connected and the loop space is simple), the obstructions to doing so lie in \(H^{n+1}(BG; \pi_n(\Omega \widetilde{\operatorname{BGL}_{\otimes}(X; {\mathbb{F}})^+})) = H^{n+1}(BG; \pi_{n+1}(\operatorname{BGL}_{\otimes}(X; {\mathbb{F}})^+))\). Since \(G\) is rationally acyclic, the cohomology groups of \(G\) with rational coefficients are zero in degree above zero. Lemma 23 then concludes the proof. ◻

Proof. The case for \(\varphi\) being trivially stably linearizable reduces to the case of being linearizable, as we could just replace \(\varphi\) by \(\varphi \otimes c_{id}\) without changing the induced map \(BG \to B\mathcal{Q}(X)\).

Now suppose \(\varphi: G \to \mathcal{Q}(X, q)\) is linearizable. Recall that two group homomorphisms \(f, g: G_1 \to G_2\) induce freely homotopic maps \(Bf, Bg: BG_1 \to BG_2\) if and only if they differ by conjugations. Thus, the map \(B\varphi: BG \to B\mathcal{Q}(X, q)\) is freely homotopic to a map \(B\varphi': BG \to B\mathcal{Q}(X, q)\) such that \(\varphi': G \to \mathcal{Q}(X, q)\) is linearly onsite. Lemma 24 then shows that \[(\varphi')^{st}: BG \xrightarrow{B\varphi'} B\mathcal{Q}(X, q) \to B\mathcal{Q}(X) \to B\mathcal{Q}(X)^+ \xrightarrow{\sim} K(\mathbf{C}(X))_1\] is null-homotopic. This shows that the map \[(\varphi)^{st}: BG \xrightarrow{B\varphi} B\mathcal{Q}(X, q) \to B\mathcal{Q}(X) \to B\mathcal{Q}(X)^+ \xrightarrow{\sim} K(\mathbf{C}(X))_1\] is freely null-homotopic. For a space \(Y\) and a connected \(H\)-space \(Z\), based-homotopy class of maps \(Y \to Z\) are the same as free homotopy classes \(Y \to Z\) (see Proposition 1.4.3 of [33], for example). Thus, \((\varphi')^{st}\) is also based null-homotopic. This concludes the proof. ◻

Proof of Theorem [thm::fin95gp95null]. Suppose \(\varphi\) is stably linearizable, then there is a linearizable \(\psi: G \to \mathcal{Q}(X, r)\) such that \((\varphi \otimes \psi)\) is linearizable. Consider the stabilization map \[(\varphi \otimes \psi)^{st}: BG \xrightarrow{B(\varphi \otimes \psi)} B\mathcal{Q}(X, qr) \to B\mathcal{Q}(X) \to B\mathcal{Q}(X)^+ \xrightarrow{\sim} K(\mathbf{C}(X))_1.\] This is homotopic to the map \[BG \xrightarrow{(\varphi^{st}, \psi^{st})} B\mathcal{Q}(X)^+ \times B\mathcal{Q}(X)^+ \xrightarrow{\mu} B\mathcal{Q}(X)^+,\] where \(\mu\) is the \(H\)-space structure on \(B\mathcal{Q}(X)^+\). In particular, in the group \([BG, B\mathcal{Q}(X)^+]_*\), \((\varphi \otimes \psi)^{st}\) represents the class \([\varphi^{st}] + [\psi^{st}]\). From Corollary 25, we know \([\psi^{st}] = 0\) in this group. Thus we have that \(0 = [\varphi^{st}] + 0\), so \(\varphi^{st}\) is also based null-homotopic.
Suppose \(\varphi\) is weakly linearizable, evidently the argument for null-homotopy only relied on the stable diagram being factorizable, and we analogously can do this modulo conjugations. Thus, \(\varphi^{st}\) is null-homotopic. ◻

Proof of Lemma 23. We first note that \(\operatorname{BGL}(X; {\mathbb{F}})^+\) is connected, so there is no issue at \(\pi_0\) or \(H_0\). Write \(\operatorname{QS}(X)\) as a poset whose objects are functions \(q: X \to {\mathbb{Z}}_{>0}\) and \(q \leq r\) if \(q(x)\) divides \(r(x)\) for all \(x \in X\). In this case, we write \(r = q \cdot q'\). For a natural number \(n \in {\mathbb{Z}}_{>0}\), we also write \(n\) to denote the constant function \(n: X \to \{n\}\).
We note that each \(q \in \operatorname{QS}(X)\) corresponds to the component \(B\mathcal{C}_{q}\) corresponding to \(q \in \pi_0(B\mathcal{C})\). We write \([q]\) to denote the corresponding element in \(H_0(K\mathcal{C})\). Furthermore, for \(q \leq r\) with \(r = q \cdot q'\), there is a natural map \(B\mathcal{C}_q \to B\mathcal{C}_r\) given by multiplication by \(q'\).
The set-up above enables us to consider the following strict filtered colimit \[L \mathrel{\vcenter{:}}= \operatorname{colim}_{q \in \operatorname{QS}(X)} B\mathcal{C}_{q}.\] By Corollary 27, \(\operatorname{BGL}(X; {\mathbb{F}})^+\) is the identity component of \(K(\mathcal{D})\). Since \(K(\mathcal{D})\) is the group completion of \(B\mathcal{D}\), the axiomatic definition of localization for group completion implies that for any commutative ring \(R\), \[H_*(K(\mathcal{D}); R) \cong \pi_0(B\mathcal{D})^{-1} H_*(B\mathcal{D}; R).\] In particular, the homology at the identity component corresponds to applying the localization (which is equivalently a colimit) at the homology of the identity component of \(B\mathcal{D}\). In particular, this shows \[H_*(\operatorname{BGL}(X; {\mathbb{F}})^+; R) \cong H_*(K(\mathcal{D})_1; R) \cong H_*(L; R) \cong \operatorname{colim}_{q \in \operatorname{QS}(X)} H_*(B\mathcal{C}_{q}; R).\] For a prime \(p\), we wish to show that \(\tilde{H}_*(L; {\mathbb{F}}_p) = 0\). The universal coefficient theorem for homology will then imply the multiplication by \(p\), in the sense of adding an element with itself \(p-1\) times, is an isomorphism on \(\tilde{H}_*(L; {\mathbb{Z}})\). If this holds for all primes \(p\), then \(\tilde{H}_*(L; {\mathbb{Z}})\) is rational. It then follows from Lemma 28 that \(\operatorname{BGL}(X; {\mathbb{F}})^+\) has rational homotopy groups.
Since \(B(\mathcal{C})\) is an \({\mathbb{E}}_{\infty}\)-ring space, the homology group \(H_*(B\mathcal{C}; {\mathbb{F}}_p)\) has two ring structures under the H-space structure given by \(\oplus\) and the H-space structure given by \(\otimes\). We denote products with respect to \(\oplus\) as \(*\) and products with respect to \(\otimes\) as \(\bullet\). We now make a subclaim that for any \(x \in H_n(B\mathcal{C}_q; {\mathbb{F}}_p)\) for \(n > 0\) and \(q \in \operatorname{QS}(X)\), for \(t\) large enough, \(x \bullet [p^t] \in H_n(B\mathcal{C}_{q\cdot p^k}; {\mathbb{F}}_p)\) is \(0\). This subclaim will imply that \(\tilde{H}_*(L; {\mathbb{F}}_p) = 0\) for the following reason. Since \(\tilde{H}_*(L; {\mathbb{F}}_p) = \operatorname{colim}_{q \in \operatorname{QS}(X)} \tilde{H}_*(B\mathcal{C}_q; {\mathbb{F}}_p)\) is a filtered colimit of abelian groups, we just need to show that any element \(x \in \tilde{H}_*(B\mathcal{C}_q; {\mathbb{F}}_p)\) becomes \(0\) at some point in the colimit stage. By the subclaim, it suffices to find a path in the poset \(\operatorname{QS}(X)\) that goes through infinitely many \(\bullet\) by \([p]\), and this path always exists.
It remains for us to show the subclaim. We first discuss a construction that will be helpful. In general for a space \(Y\), the homology \(H_*(Y; {\mathbb{F}}_p)\) has a coproduct map \(H_*(Y; {\mathbb{F}}_p) \to H_*(Y \times Y; {\mathbb{F}}_p) \cong H_*(Y; {\mathbb{F}}_p) \otimes H_*(Y; {\mathbb{F}}_p)\) induced first by the diagonal map \(\Delta: Y \to Y \times Y\) and then the Kunneth isomorphism. We use \(\psi: H_*(Y; {\mathbb{F}}_p) \to H_*(Y; {\mathbb{F}}_p)^{\otimes p}\) to denote the \(p\)-fold iterated coproduct map. The map \(\psi\) satisfies a co-commutativity condition in the sense that \(\psi\) is invariant under any permutation \(\sigma\) on the tensors of \(H_*(Y; {\mathbb{F}}_p)^{\otimes p}\) (i.e., \(\sigma \circ \psi = \psi\)). Write \(\{x_i\}_{i \in I}\) as a basis of \(H_*(Y; {\mathbb{F}}_p)\) where each \(x_i\) has pure degree. Write \(I = (i_1, ..., i_p)\) as a multi-index such that \(\deg(x_{i_1}) + ... \deg(x_{i_p}) = k\). An element \(x \in H_k(Y; {\mathbb{F}}_p)\) will be sent to \(\psi(x) = \sum a_I x_{i_1} \otimes ... \otimes x_{i_p}\). Let \(\sigma\) be a permutation of \(p\)-elements, we note that the co-commutativity of \(\psi\) implies that \(a_{I} = a_{\sigma I}\).
Proposition II.1.5 [34] computed how \(*\) and \(\bullet\) interact with each other in the homology of any \({\mathbb{E}}_{\infty}\)-ring space. In particular, for \(x \in H_n(B\mathcal{C}_q; {\mathbb{F}}_p) \subseteq H_n(B\mathcal{C}; {\mathbb{F}}_p)\) it implies that \[x \bullet [p] = x\bullet (\underbrace{[1] * ... * [1]}_{\text{p times}}) = \sum a_I x_{i_1} * ... * x_{i_p}.\] Since \(*\) is commutative, we can group together any summand that differs from each other by a permutation \(\sigma\) of \(p\)-elements. There is then \(\frac{p!}{n_1! ... n_k!}\) many copies of the same summand where \(n_1, ..., n_k\) are the number of the same elements in \(I\). In particular, \(\frac{p!}{n_1! ... n_k!}\) is divisible by \(p\) unless \(x_{i_1} = ... = x_{i_p} = z_j\) for some basis element \(z_j\). Thus, we have that \[x \bullet [p] = \sum z_j^{[p]},\] where \(z_j^{[p]}\) denotes \(p\)-th power of \(z_j\) under \(*\). Now \(H_*(B(\mathcal{C}); {\mathbb{F}}_p)\) is an \({\mathbb{F}}_p\)-algebra under \(*\), so we have that \[x \bullet [p] = (\sum z_j)^{[p]}.\] Write \(y = \sum z_j\), we now have that \(x \bullet [p] = y^{[p]}\). Now observe that if \(p\) does not divide \(\deg(x) = n\), then \(y = 0\) (this is because \(z_j^{[p]}\) has degree \(p \deg(z)\), which is equal to \(q > 0\)). Otherwise, we multiply \(x\) by \([p^k] = [p] \bullet ... \bullet [p]\) until \(\frac{\deg(x)}{p^k} = \frac{n}{p^k}\) is no longer an integer. This shows that \(x \bullet [p^k] = 0\) for \(k\) sufficiently large. ◻

Proof of Theorem [thm::converse95on95PGL]. Suppose \(\varphi: G \to \mathcal{Q}(X, q)\) restricts to \(\prod_{x \in X} \operatorname{PGL}_{q(x)}({\mathbb{F}})\) and induces a null-homotopic map \(BG \to \operatorname{BPGL}(X; {\mathbb{F}})^+\). The surjection \(\operatorname{GL}_{\otimes}(X; {\mathbb{F}}) \to \operatorname{PGL}(X; {\mathbb{F}})\) has central kernel and satisfies the main theorem of [21], so the induced map \(\operatorname{BGL}_{\otimes}(X; {\mathbb{F}})^+ \to \operatorname{BPGL}(X; {\mathbb{F}})^+\) is a fibration. The homotopy lifting property of fibrations allows us to lift the null-homotopy \(BG \to \operatorname{BPGL}(X; {\mathbb{F}})^+\) to a map \(BG \to \operatorname{BGL}_{\otimes}(X; {\mathbb{F}})^+\). We now have a square \[\begin{tikzcd} BG && \\ & {\operatorname{BGL}_{\otimes}(X; \mathbb{F})} & {\operatorname{BPGL}(X; \mathbb{F})} \\ & {\operatorname{BGL}_{\otimes}(X; \mathbb{F})^+} & {\operatorname{BPGL}(X; \mathbb{F})^+} \arrow["{B\varphi}", from=1-1, to=2-3] \arrow["{\text{lift}}"', dashed, from=1-1, to=3-2] \arrow[from=2-2, to=2-3] \arrow[from=2-2, to=3-2] \arrow[from=2-3, to=3-3] \arrow["{\text{fibration}}"', from=3-2, to=3-3] \end{tikzcd}\] By Lemma 1 below, the bottom square is in fact a pullback square, so we have a map \(BG \to \operatorname{BGL}_{\otimes}(X; {\mathbb{F}})\) given by the pullback. This shows \(\varphi\) is weakly linearizable. When \(G\) is finite, the induced map \(G \to \operatorname{GL}_{\otimes}(X; {\mathbb{F}})\) must land in some finite stage. This shows \(\varphi: G \to \mathcal{Q}(*, q)\) is stably linearizable. ◻

Proof. For a fibration \(F \to E \to B\) where:

• \(B\) is path-connected.

• \(\pi_1(B)\) acts trivially on \(\pi_n(F)\).

• \(F\) is connected simple with \(\pi_1(F)\).

(call these conditions \((\dagger)\)), there is a canonical class in \(H^2(B; \pi_1(F))\) given as follows. Consider the cohomological Serre spectral sequence of the fibration \(F \to E \to B\), there is a \(d_2\) differential arising \[d_2: E^{0,1}_2 = H^0(B; H^1(F; \pi_1F)) \cong H^1(F; \pi_1 F) \cong \operatorname{Hom}(\pi_1 F; \pi_1 F)\]\[\to E^{2,0}_2 = H^2(B; H^0(F; \pi_1 F)) = H^2(B; \pi_1 F).\] The image of the identity map \(\pi_1 F \to \pi_1 F\) under \(d_2\) defines a canonical class in \(H^2(B; \pi_1 F)\). Note that when we consider the fibration \(BH \to BG \to BG/H\), which satisfies \((\dagger)\), this class is the cohomology class \(\beta\) classifying the central extension \(H \to G \to G/H\).
The induced fibration \(BH \to BG^+ \to BG/H^+\) also satisfies the condition \((\dagger)\) above, and hence has a cohomology class \(\alpha\). Pullbacks respect this cohomology class. Note that a pull-back \[\begin{tikzcd} P & {BG/H} \\ {BG^+} & {BG/H^+} \arrow[from=1-1, to=1-2] \arrow[from=1-1, to=2-1] \arrow[from=1-2, to=2-2] \arrow[from=2-1, to=2-2] \end{tikzcd}\] implies that there is a fibration \(BH \to P \to BG/H\), which must imply \(P \cong B\pi_1 P\) from a long exact sequence computation. It suffices to show the extension class \(\gamma\) classifying \(H \to \pi_1(P) \to G/H\) is the same as the class \(\beta\) for \(H \to G \to G/H\).
Let \(i: BG/H \to BG/H^+\) denote the plus-construction map. By naturality we have that \(i^*(\alpha) = \gamma\). On the other hand, clearly the Serre spectral sequence for the fibration \(BH \to BG^+ \to BG/H^+\) is isomorphic to the Serre spectral sequence for the fibration \(BH \to BG \to BG/H\) via the plus-construction map. This isomorphism would send \(i^*(\alpha) = \beta\) in the \(E_2\)-page, in particular. Thus, we have that \(\beta = \gamma\), and hence \(P = BG\) and the initial square is a pullback square. ◻

Proof. We modify the construction to Definition 26 to consider a groupoid \(\mathcal{C}^*\) whose objects are assignments to each \(x \in X\) a vector space \({\mathbb{C}}^{f(x)}\) and the automorphisms are given by \(\prod_{x \in X} U_{f(x)}\), with operations given by direct sum and tensor product. Proposition 1.5 of [35] still applies to identify \(K(\mathcal{C}^*)\) as \(\operatorname{BU}_{\otimes}(X)^+\). A nearly identical argument as in the proof of Lemma 23 shows that \(K(\mathcal{C}^*)\) is rational. Thus, the rest of the argument in Theorem [thm::fin95gp95null] proceeds similarly. ◻

Proof. The exact sequence \(U(1) \to U_{\otimes} \to PU\) induces a surjective map \(U_{\otimes}(X) \to \operatorname{PU}(X)\) with central kernel. This satisfies the criterion in [21] and hence gives a fibration \(BU_{\otimes}(X)^+ \to B\operatorname{PU}(X)^+\). The proof then proceeds similarly as in Theorem [thm::converse95on95PGL] with the relevant square still being a pullback, due to Lemma 1. ◻

Proof. We first show that \(\widetilde{C}\) is perfect. The five-term exact sequence for the central extension \(1\to A_1\to \widetilde{C}\to C\to 1\) is \[H_2(\widetilde{C};\mathbb{Z})\longrightarrow H_2(C;\mathbb{Z})\longrightarrow A_1\longrightarrow H_1(\widetilde{C};\mathbb{Z})\longrightarrow H_1(C;\mathbb{Z})\longrightarrow 0.\] Since \(C\) is perfect, \(H_1(C;\mathbb{Z})=0\). Moreover, by construction of the universal central extension, the map \(H_2(C;\mathbb{Z})\to A_1\) is the identity map on \(A_1=H_2(C;\mathbb{Z})\). Exactness therefore gives \(H_1(\widetilde{C};\mathbb{Z})=0\).

It remains to prove that \(H_2(\widetilde{C};\mathbb{Z})=0\). Let \(1\to A\to E\to \widetilde{C}\to 1\) be any central extension. Since \(\widetilde{C}\) is perfect, the composite \(E\to \widetilde{C}\to C\) is again a central extension of \(C\). Indeed, the preimage of \(A_1\subset \widetilde{C}\) is central in \(E\); commutators with such lifts define homomorphisms \(\widetilde{C}\to A\), and these vanish because \(\widetilde{C}\) is perfect.

By the universal property above, there is a unique map \(\widetilde{C}\to E\) over \(C\). Composing with \(E\to \widetilde{C}\) gives a map \(\widetilde{C}\to \widetilde{C}\) over \(C\), hence the identity by the same universal property. Thus, the central extension \(E\) of \(\widetilde{C}\) splits.

Since \(\widetilde{C}\) is perfect, central extensions of \(\widetilde{C}\) by an abelian group \(A\) are classified by \(H^2(B\widetilde{C};A)\cong \operatorname{Hom}(H_2(\widetilde{C};\mathbb{Z}),A)\). Taking \(A=H_2(\widetilde{C};\mathbb{Z})\), the identity map must correspond to a split extension, hence to the zero class. Therefore \(H_2(\widetilde{C};\mathbb{Z})=0\). ◻

Proof. The homotopy limit \(X_\infty\) is acyclic, i.e., its reduced integral homology vanishes in all degrees. Moreover, the map \(X_\infty\to BC\) is surjective on fundamental groups. Therefore the homotopy cofiber of \(X_\infty\to BC\) is simply connected, and the induced map from \(BQ\) to this homotopy cofiber is a homology equivalence. By the characterization of the plus construction, this homotopy cofiber is equivalent to \(BC^+\). In other words, \(X_\infty\) is the homotopy fiber of \(BC\rightarrow BC^+.\) Since the square \[\begin{tikzcd} BQ \arrow[r] \arrow[d] & BQ^+ \arrow[d] \\ BC \arrow[r] & BC^+ \end{tikzcd}\] is cartesian and the vertical maps are covering maps, the homotopy fibers of the two horizontal maps are naturally equivalent, after choosing compatible basepoints. Since \(X_\infty\) is the homotopy fiber of \(BC\to BC^+\), it follows that \(X_\infty\) is also the homotopy fiber of \(BQ\to BQ^+\). ◻

Proof of Lemma 2. Consider the Serre spectral sequence of the fibration \[K(A_n,n)\longrightarrow X_{n+1}\longrightarrow X_n .\] with \[E^2_{p,q}\cong H_p(X_n;H_q(K(A_n,n);\mathbb{Z})).\]

The fiber \(K(A_n,n)\) is \((n-1)\)-connected. In low degrees, its homology is \[H_q(K(A_n,n);\mathbb{Z})\cong \begin{cases} \mathbb{Z}, & q=0,\\ 0, & 0<q<n,\\ A_n, & q=n,\\ 0, & q=n+1. \end{cases}\] On the other hand, \(H_p(X_n;\mathbb{Z})=0\) for \(0<p\leq n\).

It follows that, in total degree at most \(n+1\), the only possibly nonzero terms are \(E^2_{0,0}\cong \mathbb{Z}\), \(E^2_{n+1,0}\cong H_{n+1}(X_n;\mathbb{Z})\), and \(E^2_{0,n}=H_0(X_n, H_n(K(A_n,n);\mathbb{Z}))\).

The differential has bidegree \[d^r:E^r_{p,q}\longrightarrow E^r_{p-r,q+r-1}.\] The term \(E^r_{n+1,0}\) receives no differentials, since an incoming differential would have source \(E^r_{n+1+r,1-r}\), which vanishes for \(r\geq 2\).

For \(2\leq r\leq n\), the outgoing differential \[d^r:E^r_{n+1,0}\longrightarrow E^r_{n+1-r,r-1}\] has target in a row \(0<r-1<n\), and this row vanishes because \(H_{r-1}(K(A_n,n);\mathbb{Z})\) vanishes. Therefore the first possible nonzero differential out of \(E_{n+1,0}\) is \(d^{n+1}.\) Under the identifications above, this is a homomorphism \[d^{n+1}:H_{n+1}(X_n;\mathbb{Z})\to H_0(X_n, H_n(K(A_n,n);\mathbb{Z}))\cong A_n.\]

We now identify this differential. By naturality of the Serre spectral sequence the transgression \(d^{n+1}\) is evaluation against the classifying class \(\iota_n\in H^{n+1}(X_n;A_n)\). Therefore the differential is an isomorphism.

Consequently, both terms \(E^{n+1}_{n+1,0}\) and \(E^{n+1}_{0,n}\) are killed by this differential. It follows that \(\widetilde{H}_i(X_{n+1};\mathbb{Z})=0\) for \(i\leq n+1\). Therefore \(X_{n+1}\) is homologically \((n+1)\)-connected. ◻

Proof. Let \[p_i:X_{i+1}\longrightarrow X_i\] denote the \(i\)th stage of the tower. By the construction of the tower, the class \[\iota_i\in H^{i+1}(X_i;A_i)\] is the universal obstruction to lifting a map into \(X_i\) through \(p_i\). More precisely, if \(f:Y\to X_i\) is a map, then the primary obstruction to the existence of a lift \(\widetilde{f}:Y\to X_{i+1}\) with \(p_i\circ \widetilde{f}\cong f\) is the class \[f^*\iota_i\in H^{i+1}(Y;A_i).\] The vanishing of this class is equivalent to the existence of such a lift.

We apply this with \(Y=BG\) and argue by induction. If all obstruction classes vanish, the inductive argument produces a compatible system of lifts \({\widetilde{\varphi}_i:BG\to X_i}\). These maps determine a map into the homotopy inverse limit \[\widetilde{\varphi}_\infty:BG\longrightarrow X_\infty.\]

Finally, by construction, \(X_\infty\) is the homotopy fiber of the map \[BQ\longrightarrow BQ^+.\] Therefore a map \(BG\to BQ\) admits a lift to \(X_\infty\) if and only if its composite \[BG\longrightarrow BQ\longrightarrow BQ^+\] is null-homotopic. This proves the final assertion. ◻

Proof. First, \(A_0=\mathcal{Q}(\mathbb{Z}^n)/ \mathcal{C}(\mathbb{Z}^n)=\pi_0 \mathbf{Q}(\mathbb{Z}^n; \mathbb{F})\). Second, \(A_1=H_2(\mathcal{C}(\mathbb{Z}^n); \mathbb{Z})=\pi_1 \mathbf{Q}(\mathbb{Z}^n; \mathbb{F})\) according to Corollary C of [5]. For \(n\geq 2\), \(A_n=\pi_{n}X_\infty\) by Dror’s tower. Since \(BQ\) is a classifying space, it is a \(K(Q,1)\). Thus \[\pi_i(BQ)=0\] for \(i\geq 2\), while \(\pi_1(BQ)\cong Q\). By Theorem 36, there is a long exact sequence for \(n\geq 3\), \[0=\pi_n(BQ)\longrightarrow \pi_n(BQ^+) \longrightarrow \pi_{n-1}(X_\infty) \longrightarrow \pi_{n-1}(BQ)=0.\] Hence the connecting homomorphism is an isomorphism \[\pi_n(BQ^+)\cong \pi_{n-1}(X_\infty).\] We obtain \[\pi_{n+1}(BQ^+)\cong A_{n}, \qquad n\geq 2.\] The result is then implied by Theorem D and E of [5]. ◻

Proof. The proof is identical, using the unitary analogs of the same results from [5]. ◻

Proof. It follows from Theorem [thm::fin95gp95null], Proposition 37, Lemmas 3 and 4. ◻

Proof. The obstruction class associated to \(A_i\) lies in \(H^{i+1}(BG;A_i)\). By the second columns of Table ¿tbl:table95q95z? (which are derived from Table 1 of [5]), for \(0\leq i\leq n\) the groups \(A_i\) are identified with \[\mathcal{Q}(\mathbb{Z}^{n-i})/\mathcal{C}(\mathbb{Z}^{n-i}).\] The next non-trivial contribution is the \({\mathbb{Q}}/\mathbb{Z}\) summand in \(A_{n+2}\). All remaining terms, including \(K_2(\mathbb{C}) \otimes \mathbb{Q}\), are rational vector spaces. Since \(G\) is finite (or rationally acyclic), its rational cohomology vanishes in positive degree, and hence these rational summands do not give nonzero obstruction classes. ◻

Proof. The proof is identical to the algebraic counterpart over \(\mathbb{C}\) using the unitary column of Table ¿tbl:table95q95z? instead. The unitary column is derived from a later computation in Proposition 46. ◻

Proof of Theorem 43. Since \(\mathbf{Q}({\mathbb{Z}}^1; {\mathbb{C}})\) is the (based) loop space of \(K(\mathbf{C}({\mathbb{Z}}^1; {\mathbb{C}}))\) and \(\Omega\) commutes with products, it suffices to show that \(K(\mathbf{C}({\mathbb{Z}}^1; {\mathbb{C}}))\) is equivalent to a product of Eilenberg-MacLane spaces. Since \(K(\mathbf{C}({\mathbb{Z}}^1; {\mathbb{C}}))\) is Segal’s K-theory of a symmetric monoidal category, it is in particular the zeroth space of a connective spectrum \(F\) (see Appendix of [31]).

Theorem D of [5] shows us that \(K(\mathbf{C}({\mathbb{Z}}^1; {\mathbb{C}})) \cong K(\mathbf{C}({\mathbb{Z}}^1; {\mathbb{C}}))_1\). We are reduced to showing \(K(\mathbf{C}({\mathbb{Z}}^1; {\mathbb{C}}))_1\) is a product of Eilenberg-MacLane spaces, which we will do by showing \(F\) has vanishing stable \(k\)-invariants and is hence a product of Eilenberg-MacLane spaces.

Indeed, the homotopy groups of \(F\) are the homotopy groups of \(K(\mathbf{C}({\mathbb{Z}}^1; {\mathbb{C}}))_1\), which are given in Table ¿tbl:table95q95z? of this paper (and also Section 6 of [5]): \[\begin{array}{c|ccccc} i & 0 & 1 & 2 & 3 & \geq 3 \\ \hline \pi_i(F) & 0 & {\mathbb{Q}}_{> 0} & 0 & \frac{{\mathbb{Q}}}{{\mathbb{Z}}} \oplus (K_2(\mathbb{C}) \otimes {\mathbb{Q}}) & \text{rational vector space} \end{array}\] The first possibly non-trivial \(k\)-invariant occurs between \(\pi_1\) and \(\pi_{3}\) as a map \(\tau_{\leq 3-1} F \cong \Sigma^1 H{\mathbb{Q}}_{>0} \to \Sigma^4 H(\frac{{\mathbb{Q}}}{{\mathbb{Z}}} \oplus K_2(\mathbb{C} \otimes {\mathbb{Q}}))\) which represents an element in \[[\Sigma^1 H{\mathbb{Q}}_{>0}, \Sigma^4 H(\frac{{\mathbb{Q}}}{{\mathbb{Z}}} \oplus K_2(\mathbb{C} \otimes {\mathbb{Q}}))] \cong [H{\mathbb{Q}}_{>0}, \Sigma^3 H(\frac{{\mathbb{Q}}}{{\mathbb{Z}}} \oplus K_2(\mathbb{C} \otimes {\mathbb{Q}}))],\] which is \(H^3(H{\mathbb{Q}}_{>0}; \frac{{\mathbb{Q}}}{{\mathbb{Z}}} \oplus K_2(\mathbb{C} \otimes {\mathbb{Q}})) \cong H^3(H{\mathbb{Q}}_{>0}; \frac{{\mathbb{Q}}}{{\mathbb{Z}}}) \oplus H^3(H{\mathbb{Q}}_{>0}; K_2(\mathbb{C}) \otimes {\mathbb{Q}})\). Observe that \({\mathbb{Q}}_{>0}\) is isomorphic to a countably infinite direct sum of integers, indexed over the prime numbers, so we can without loss write \({\mathbb{Q}}_{>0} \cong \bigoplus_{{\mathbb{N}}} {\mathbb{Z}}\). We can then rewrite \[H^3(H{\mathbb{Q}}_{>0}; \frac{{\mathbb{Q}}}{{\mathbb{Z}}}) \cong H^3(H\bigoplus_{{\mathbb{N}}} {\mathbb{Z}}; \frac{{\mathbb{Q}}}{{\mathbb{Z}}}) \cong \prod_{{\mathbb{N}}} H^3(H{\mathbb{Z}}; \frac{{\mathbb{Q}}}{{\mathbb{Z}}}),\] \[H^3(H{\mathbb{Q}}_{>0}; K_2(\mathbb{C}) \otimes {\mathbb{Q}}) \cong H^3(H\bigoplus_{{\mathbb{N}}} {\mathbb{Z}}; K_2(\mathbb{C}) \otimes {\mathbb{Q}}) \cong \prod_{{\mathbb{N}}} H^3(H{\mathbb{Z}}; K_2(\mathbb{C}) \otimes {\mathbb{Q}}).\] Now we claim that both \(H^3(H{\mathbb{Z}}; \frac{{\mathbb{Q}}}{{\mathbb{Z}}})\) and \(H^3(H{\mathbb{Z}}; K_2(\mathbb{C}) \otimes {\mathbb{Q}})\) are \(0\), which would show this \(k\)-invariant is trivial. We defer the proof of the first case to the technical Lemma 5. For the second case, since \(K_2(\mathbb{C}) \otimes {\mathbb{Q}}\) is a rational vector space, it suffices to check that \(H^3(H{\mathbb{Z}}; {\mathbb{Q}}) = 0\). In fact, we will show that \(H^k(H{\mathbb{Z}}; {\mathbb{Q}}) = 0\) for all \(k > 0\). The cohomology \(H^k(H{\mathbb{Z}}; {\mathbb{Q}})\) can be computed as the colimit of unstable cohomology groups \(\operatorname{colim}_n H^{n+k}(K({\mathbb{Z}}, n); {\mathbb{Q}})\). For \(n > 1\), \(K({\mathbb{Z}}, n)\) is simply connected and hence the rational equivalence \(K({\mathbb{Z}}, n) \cong_{\mathbb{Q}} K({\mathbb{Q}}, n)\) shows that \(H^{n+k}(K({\mathbb{Z}}, n); {\mathbb{Q}}) \cong H^{n+k}(K({\mathbb{Q}}, n); {\mathbb{Q}})\). For \(n\) odd, \(S^n\) is rationally equivalent to \(K({\mathbb{Q}}, n)\), so \(H^{n+k}(K({\mathbb{Q}}, n); {\mathbb{Q}}) = 0\). For \(n\) even, we have that \(\widetilde{H}_i(K({\mathbb{Q}}, n); {\mathbb{Z}}) = {\mathbb{Q}}\) for every \(i\) a positive multiple of \(n\) and is equal to \(0\) otherwise. The universal coefficient theorem then shows that \(H^{n+k}(K({\mathbb{Q}}, n); {\mathbb{Q}}) = 0\) for \(n\) sufficiently large. It follows that \(H^k(H{\mathbb{Z}}; {\mathbb{Q}}) = 0\) for \(k > 0\). Note this proof also shows that \(H^k(H{\mathbb{Q}}; {\mathbb{Q}}) = 0\) for \(k > 0\).

We have now shown that \(\tau_{\leq 3} F\) is a wedge of suspensions of Eilenberg-MacLane spectra. To show \(F\) is a wedge of suspensions of Eilenberg-MacLane spectra, we need to check the higher \(k\)-invariants. However, we see that the higher \(k\)-invariants all take 1 of the following 3 forms: (a) \({\mathbb{Q}}_{>0} \to\) , (b) , and (c) \(\frac{{\mathbb{Q}}}{{\mathbb{Z}}} \to\) rational. Case (a) and (b) will have trivial \(k\)-invariants since \(H^k(H{\mathbb{Z}}; {\mathbb{Q}}) = H^k(H{\mathbb{Q}}; {\mathbb{Q}}) = 0\) for \(k > 0\). For Case (c), the fiber sequence \(H{\mathbb{Z}}\to H{\mathbb{Q}}\to H\frac{{\mathbb{Q}}}{{\mathbb{Z}}}\) gives a long exact sequence in rational cohomology. Since \(H^k(H{\mathbb{Q}}; {\mathbb{Q}}) = 0\) for \(k > 0\), it follows that \[H^k(H\frac{{\mathbb{Q}}}{{\mathbb{Z}}}; {\mathbb{Q}}) = H^{k+1}(H{\mathbb{Z}}; {\mathbb{Q}}) = 0.\] Thus, Case (c) also gives trivial \(k\)-invariants. We conclude that \(F\) is a wedge of suspensions of Eilenberg-MacLane spectra, so its zeroth space \(K(\mathbf{C}({\mathbb{Z}}^1; \mathbb{C}))_1\) is a product of Eilenberg-MacLane spaces. ◻

Proof. Let \(\widetilde{K(\mathbf{C}({\mathbb{Z}}^2; {\mathbb{C}})))_1}\) denote the universal cover. The main theorem of [5] tells us that the homotopy groups of \(K(\mathbf{C}({\mathbb{Z}}^2; {\mathbb{C}})))_1\) above degree \(1\) can be obtained from the homotopy groups of \(K(\mathbf{C}({\mathbb{Z}}^1; {\mathbb{C}})))_1\). Since the universal cover is simply connected, we then have the following table: \[\begin{array}{c|cccccc} i & 0 & 1 & 2 & 3 & 4 & \geq 5 \\ \hline \pi_i \text{ of the universal cover} & 0 & 0 & {\mathbb{Q}}_{> 0} & 0 & \frac{{\mathbb{Q}}}{{\mathbb{Z}}} \oplus (K_2(\mathbb{C}) \otimes {\mathbb{Q}}) & \text{rational vector space} \end{array}\] We see that the \(k\)-invariant questions we encounter are in the cohomology groups as those occurring in the proof of Theorem 43, which we previously checked are zero. Thus, we have that \(\widetilde{K(\mathbf{C}({\mathbb{Z}}^2; {\mathbb{C}})))_1}\) is a product of Eilenberg-MacLane spaces. Since \(\mathbf{Q}({\mathbb{Z}}^2; {\mathbb{C}})\) is group-like, we know it splits off as \(\pi_0 \mathbf{Q}({\mathbb{Z}}^2; \mathbb{C}) \times (\mathbf{Q}({\mathbb{Z}}^2; \mathbb{C}))_{1}\). The component at identity \(\mathbf{Q}({\mathbb{Z}}^2; \mathbb{C})\) is the loop space of \(\widetilde{K(\mathbf{C}({\mathbb{Z}}^2; {\mathbb{C}})))_1}\). Thus, we have that \(\mathbf{Q}({\mathbb{Z}}^2; {\mathbb{C}})\) is a product of Eilenberg-MacLane spaces.

A similar discussion, without needing to go to the universal cover, shows the case for \({\mathbb{Z}}^0\) due to the structure of its homotopy groups from the delooping. ◻

Proof of Lemma 5. Consider the fiber sequence \(H{\mathbb{Z}}\to H{\mathbb{Q}}\to H\frac{{\mathbb{Q}}}{{\mathbb{Z}}}\). In the proof of Theorem 43, we showed that \((H{\mathbb{Q}})^k(H{\mathbb{Z}}) = 0\) for \(k > 0\), independently of the present lemma. The long exact sequence then shows that \[(H\frac{{\mathbb{Q}}}{{\mathbb{Z}}})^3(H{\mathbb{Z}}) \cong (H{\mathbb{Z}})^4(H{\mathbb{Z}})\] are the integral cohomology operations in degree \(4\). We will now show that \((H{\mathbb{Z}})^4(H{\mathbb{Z}}) = 0\). For convenience, we write \(X = H{\mathbb{Z}}\). The proof of why \((H{\mathbb{Z}})^4(H{\mathbb{Z}}) = 0\) will follow from a series of sub-statements which we prove one by one.
(a) The Steenrod algebra \(H^k(H{\mathbb{F}}_p; {\mathbb{F}}_p) = (H{\mathbb{F}}_p)^k(H{\mathbb{F}}_p)\) have explicit basis that have been computed. Here we will use the basis description of Serre-Cartan [43], [44] (see also the Sage package page for Steenrod algebras [45]). These basis show that:

• For any prime \(p\), \(H^1(H{\mathbb{F}}_p; {\mathbb{F}}_p) = {\mathbb{F}}_p\{\beta\}\), generated by what is called the Bockstein map \(\beta\), and \(\beta^2 = 0\).

• For any prime \(p > 2\), the next \(k\) for which \(H^k(H{\mathbb{F}}_p; {\mathbb{F}}_p) \neq 0\) is \(k = 2(p-1)\).

• \(H^2(H{\mathbb{F}}_2; {\mathbb{F}}_2)\) is \({\mathbb{F}}_2 \{\operatorname{Sq}^2\}\) and \(H^3(H{\mathbb{F}}_2; {\mathbb{F}}_2) = {\mathbb{F}}_2 \{\operatorname{Sq}^3, \operatorname{Sq}^2 \operatorname{Sq}^1\}\), where \(\operatorname{Sq}^1\) is identified as the Bockstein map.

(b) \(H^1(X; {\mathbb{F}}_p) = 0\). Indeed, consider the fiber sequence \(X \xrightarrow{\times p} X \to X/p = H{\mathbb{F}}_p\). This induces a long exact sequence \[... \to H^0(X; {\mathbb{F}}_p) \xrightarrow{\times p} H^0(X; {\mathbb{F}}_p) \to H^1(H{\mathbb{F}}_p; {\mathbb{F}}_p) \to H^1(X; {\mathbb{F}}_p) \xrightarrow{\times p} H^1(X; {\mathbb{F}}_p) \to ...\] Clearly both \(\times p\) maps are zero maps, as the cohomology groups have coefficients in \({\mathbb{F}}_p\). This gives an exact sequence \[0 \to H^0(X; {\mathbb{F}}_p) \xrightarrow{\beta} H^1(H{\mathbb{F}}_p; {\mathbb{F}}_p) \to H^1(X; {\mathbb{F}}_p) \to 0,\] where \(\beta\) is the Bockstein map. Since \(H^0(X; {\mathbb{F}}_p) = [H{\mathbb{Z}}; H{\mathbb{F}}_p]= \operatorname{Hom}({\mathbb{Z}}; {\mathbb{F}}_p) = {\mathbb{F}}_p\), the map \(\beta\) is an isomorphism by (a). Thus, \(H^1(X; {\mathbb{F}}_p) = 0\).
(c) \(H^2(X; {\mathbb{F}}_2) = H^3(X; {\mathbb{F}}_2) = {\mathbb{F}}_2\). For \(k \in \{2, 3\}\), running the long exact sequence in \({\mathbb{F}}_2\)-cohomology on the fiber sequence \(X \xrightarrow{\times 2} X \to X/p\), we again have an exact sequence \[0 \to H^{k-1}(X; {\mathbb{F}}_2) \xrightarrow{\beta} H^k(HX/2 = H{\mathbb{F}}_2; {\mathbb{F}}_2) \to H^k(X; {\mathbb{F}}_2) \to 0.\] We now see that \(H^k(X; {\mathbb{F}}_2)\) is the cokernel of the Bockstein map. From (a), we then know that \(H^k(X; {\mathbb{F}}_2)\) is 1-dimensional for \(k \in \{2, 3\}\).
Now we will prove \((H{\mathbb{Z}})^4(H{\mathbb{Z}}) = H^4(X; {\mathbb{Z}}) = 0\). By Lemma 6, it suffices to show that each \(p\)-primary part of \(H^4(X; {\mathbb{Z}})\) is \(0\) and the \(p\)-primary part of \(H^k(X; {\mathbb{Z}})\) is exactly \(\ker(\times p: H^k(X; {\mathbb{Z}}) \to H^k(X; {\mathbb{Z}}))\) for \(k > 0\). Consider the exact sequence \({\mathbb{Z}}\xrightarrow{\times p} {\mathbb{Z}}\to {\mathbb{F}}_p\), which induces a long exact sequence in cohomology groups of \(X\) as: \[\begin{tikzcd} {H^1(X; {\mathbb{F}}_p)=0} & {H^2(X; {\mathbb{Z}})} & {H^2(X; {\mathbb{Z}})} & {H^2(X; {\mathbb{F}}_p)} & \\ & {H^3(X; {\mathbb{Z}})} & {H^3(X; {\mathbb{Z}})} & {H^3(X; {\mathbb{F}}_p)} \\ & {H^4(X; {\mathbb{Z}})} & {H^4(X; {\mathbb{Z}})} & {H^4(X; {\mathbb{F}}_p)} & {...} \arrow[from=1-1, to=1-2] \arrow["{\times p}", from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=2-2] \arrow["{\times p}"', from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[from=2-4, to=3-2] \arrow["{\times p}"', from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \end{tikzcd}\] where \(H^1(X; {\mathbb{F}}_p) = 0\) by (b). We know that \(H^3(X; {\mathbb{F}}_p) = 0\) is the cokernel of the Bockstein \(\beta: H^2(X; {\mathbb{F}}_p) \to H^3(H{\mathbb{F}}_p; {\mathbb{F}}_p)\) and hence (by (a)) is \(0\) for \(p > 2\), so the map \(\times p: H^4(X; {\mathbb{Z}}) \to H^4(X; {\mathbb{Z}})\) is injective. Thus, \(H^4(X; {\mathbb{Z}})\) has zero \(p\)-primary parts for \(p > 2\). When \(p = 2\), (a) implies the diagram becomes \[\begin{tikzcd} 0 & {H^2(X; {\mathbb{Z}})} & {H^2(X; {\mathbb{Z}})} & {{\mathbb{F}}_2} & \\ & {H^3(X; {\mathbb{Z}})} & {H^3(X; {\mathbb{Z}})} & {{\mathbb{F}}_2} \\ & {H^4(X; {\mathbb{Z}})} & {H^4(X; {\mathbb{Z}})} & {H^4(X; {\mathbb{F}}_2)} & {...} \arrow[from=1-1, to=1-2] \arrow["{\times 2}", from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=2-2] \arrow["{\times 2}"', from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[from=2-4, to=3-2] \arrow["{\times 2}"', from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \end{tikzcd}.\] Since the map \(\times p: H^2(X; {\mathbb{Z}}) \to H^2(X; {\mathbb{Z}})\) is injective for any prime, Lemma 6 implies \(H^2(X; {\mathbb{Z}}) = 0\). The exact sequence now reduces to \[\begin{tikzcd} 0 & {{\mathbb{F}}_2} & {{\mathbb{F}}_2} & {{\mathbb{F}}_2} & {{\mathbb{F}}_2} & \\ && {H^4(X; {\mathbb{Z}})} & {H^4(X; {\mathbb{Z}})} & {H^4(X; {\mathbb{F}}_2)} & {...} \arrow[from=1-1, to=1-2] \arrow["\cong", from=1-2, to=1-3] \arrow["{\times 2}", from=1-3, to=1-4] \arrow["\phi", from=1-4, to=1-5] \arrow["\delta", from=1-5, to=2-3] \arrow["{\times 2}"', from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=2-5, to=2-6] \end{tikzcd}.\] Since \(\ker(\phi) = \mathop{\mathrm{im}}(\times 2) = 0\), the map \(\phi: {\mathbb{F}}_2 \to {\mathbb{F}}_2\) is injective and hence an isomorphism. Thus, \(\ker(\delta) = \operatorname{im}(\phi) = {\mathbb{F}}_2\), and \(\delta\) is the zero map. This shows that the map \(\times 2: H^4(X; {\mathbb{Z}}) \to H^4(X; {\mathbb{Z}})\) is injective, so \(H^4(X; {\mathbb{Z}})\) has no non-trivial 2-primary parts.

Thus, \(H^4(X; {\mathbb{Z}})\) has no non-trivial \(p\)-primary parts for any prime \(p\). Since \(H^4(X; {\mathbb{Z}})\) is finite by Lemma 6, this shows that \((H{\mathbb{Z}})^4(H{\mathbb{Z}}) = H^4(X; {\mathbb{Z}}) = 0\). ◻

Proof. The case of \(\pi_0\) had been discussed right before this proposition. For \(\pi_1\), Bass’s definition of \(K_1\) (see Proposition 1 of [40]) tells us that \(K_1(\mathcal{PU}_{\otimes}) = \operatorname{colim}_{n \in ({\mathbb{Z}}_{> 0}, |)} (PU(n)^{\delta})^{ab}\). The main theorem of [46] shows that each \(PU(n)^{\delta}\) is perfect, so the colimit becomes \(0\).

For the higher homotopy groups, we will construct a fibration sequence to compute them. Let \(U_{\otimes} = \operatorname{colim}_{({\mathbb{Z}}_{>0}, |)} U(n)\) and \(PU = \operatorname{colim}_{({\mathbb{Z}}_{>0}, |)} PU(n)\), where the colimit is taken over the divisibility poset and an arrow \(d \xrightarrow{\times k} n\) is sent to the Kronecker product \(- \otimes I_k\).

The short exact sequences \(U(1) \to U(n) \to PU(n)\) assemble to give a short exact sequence \(U(1) \to U_{\otimes} \to PU\), which induces a fibration \[\label{eq::unitary95fibration} BU(1)^{\delta} \to BU_{\otimes}^{\delta} \to BPU^{\delta}.\tag{5}\] For a fibration \(F \to E \to B\) in general, the main theorem of [21] characterizes when \(F^+ \to E^+ \to B^+\) is a fibration: (a) \(\pi_1(E) \to \pi_1(B)\) sends the maximal perfect normal subgroup \(\mathcal{P}(\pi_1(E)) \subseteq \pi_1(E)\) surjectively onto the maximal perfect normal subgroup \(\mathcal{P}(\pi_1(B)) \subseteq \pi_1(B)\), (b) \(\mathcal{P}(\pi_1(B))\) acts trivially on \(\pi_*(F^+)\).
The sequence in (5 ) satisfies condition (b) as \(U(1)\) is central in \(U_{\otimes}\) and the plus construction of \(BU(1)^{\delta}\) is itself. Condition (a) is also satisfied as the maximal perfect normal subgroup of \(U_{\otimes}\) is \(SU_{\otimes}\), which is sent surjectively onto \(PU^{\delta}\), which is itself perfect because each \(PU(n)^{\delta}\) is perfect. Note this is similar to Proposition 6.2 of [5]. Thus we have that \[\label{eq::unitary95fibration95plus} BU(1)^{\delta} \to (BU_{\otimes}^{\delta})^+ \to (BPU^{\delta})^+\tag{6}\] is a fibration. Furthermore, Proposition 3 of [40] implies that we can identify \((BU^{\delta}_{\otimes})^+ = K(\mathcal{U_{\otimes}})_1\) and \((BPU^{\delta})^+ = K(\mathcal{PU}_{\otimes})_1\). Thus, we now have \[\label{eq::unitary95k95theory95fib} BU(1)^{\delta} \to K(\mathcal{U}_{\otimes})_1 \to K(\mathcal{PU}_{\otimes})_1\tag{7}\] As \(\pi_*(BU(1)^{\delta})\) is concentrated in degree \(1\), the long exact sequence in homotopy groups show that \(\pi_{i}(K(\mathcal{U}_{\otimes})_1) \cong \pi_i(K(\mathcal{PU}_{\otimes})_1)\) for \(i \geq 3\). A similar argument as to that of Lemma 23 would show that the homotopy groups of \(K(\mathcal{U_{\otimes}})_1\) are all rational. In fact, one can use Theorem V.5.3 of [38] directly here to show that \(K_i(\mathcal{U}_{\otimes}) = K_i(\mathcal{U}_{\oplus}) \otimes {\mathbb{Q}}\) for \(i > 0\). Regardless, we have that \(\pi_i(K(\mathcal{U}_{\otimes})_1)\) is rational for \(i \geq 3\).
Finally for \(i = 2\), we investigate the long exact sequence: \[0 \to \pi_2(K(\mathcal{U}_{\otimes})_1) \to \pi_2(K(\mathcal{PU}_{\otimes})_1) \to U(1) \xrightarrow{\phi} \pi_1(K(\mathcal{U}_{\otimes})_1) \to 0.\] Since \(\pi_1(K(\mathcal{U}_{\otimes})_1)\) is rational, the universal property of rationalization shows \(\phi\) factors through the rationalization \(U(1) \cong {\mathbb{R}}/{\mathbb{Z}}\to U(1) \otimes {\mathbb{Q}}\cong ({\mathbb{R}}/{\mathbb{Z}})/({\mathbb{Q}}/{\mathbb{Z}})\), which has kernel being the roots of unity \(\frac{{\mathbb{Q}}}{{\mathbb{Z}}}\). Thus, \(\frac{{\mathbb{Q}}}{{\mathbb{Z}}} \subseteq \ker(\phi)\). Since \(\frac{{\mathbb{Q}}}{{\mathbb{Z}}}\) is injective, we can decompose \(\ker(\phi) = \frac{{\mathbb{Q}}}{{\mathbb{Z}}} \oplus \text{rational group}\). We also know the exact sequence splits off as \[0 \to \pi_2(K(\mathcal{U}_{\otimes})_1) \to \pi_2(K(\mathcal{PU}_{\otimes})_1) \to \ker(\phi) \to 0.\] Since \(\pi_2(K(\mathcal{U}_{\otimes})_1)\) is rational and hence injective, we have that \(\pi_2(K(\mathcal{PU}_{\otimes})_1) \cong \frac{{\mathbb{Q}}}{{\mathbb{Z}}} \oplus \text{rational group}\).
Using the fact that \(K_i(\mathcal{U}_{\otimes}) = K_i(\mathcal{U}_{\oplus}) \otimes {\mathbb{Q}}\) for \(i > 0\), we can obtain the stronger second statement in the proposition. The split inclusion \(\frac{{\mathbb{Q}}}{{\mathbb{Z}}} \subset \frac{{\mathbb{R}}}{{\mathbb{Z}}}\) gives \(\frac{{\mathbb{R}}}{{\mathbb{Z}}} = \frac{{\mathbb{Q}}}{{\mathbb{Z}}} \oplus V\) where \(V\) is a rational vector space. It suffices to show we can extract \(V\) out of \(\pi_2(K(\mathcal{U}_{\otimes})_1)\). Now we know that \(\pi_2(K(\mathcal{U}_{\otimes})_1) = \pi_2(K(\mathcal{U}_{\oplus})_1) \otimes {\mathbb{Q}}\). Furthermore, Proposition 3 and Theorem 4 of [40] tells us that \[\pi_2(K(\mathcal{U}_{\oplus})_1) = H_2(BSU^{\delta}; {\mathbb{Z}}),\] where \(\operatorname{SU} = \operatorname{colim}_{({\mathbb{Z}}, \leq)} \operatorname{SU}(n)\) is the infinite special unitary group. [47] shows that \(H_2(BSU^{\delta}; {\mathbb{Z}}) \cong H_2(SU(n); {\mathbb{Z}})\) for a finite \(n\) sufficiently large (see also Section 4 of [48]). [49] shows that for any non-trivial and connected Lie group \(G\), \(H_2(BG^{\delta}; {\mathbb{Z}})\) surjects onto an uncountable rational vector space. Thus, \(H_2(BSU^{\delta}; {\mathbb{Z}}) = H_2(BSU(n)^{\delta})\) surjects onto an uncountable rational vector space. This shows that \(\pi_2(K(\mathcal{U}_{\otimes})_1)\) contains an uncountable rational vector space, so we can split \(V\) non-canonically off. This shows that \(K_2(\mathcal{PU}_{\otimes}) \cong U(1) \oplus (\text{rational group})\). ◻

Proof of Theorem 49. Due to the description of homotopy groups obtained in Proposition 46, the unitary delooping result in Theorem 5.10 of [5], the proof of Theorem 43 and Corollary 44 adapts similarly to show that \(K(\mathbf{C}^*({\mathbb{Z}}^0)), K(\mathbf{C}^*({\mathbb{Z}}^1))\), and the universal cover \(\widetilde{K(\mathbf{C}^*({\mathbb{Z}}^2))_1}\) are all products of Eilenberg-MacLane spaces. Now \(\pi_1(K(\mathbf{C}^*({\mathbb{Z}}^2)))\) is the unitary QCA classification group of a plane, which had been shown to be \(0\) (see Section 3 of [10] or Appendix of [27]), so \(K(\mathbf{C}^*({\mathbb{Z}}^2))\) is its own universal cover. This concludes the proof. ◻

Proof. Write \(X = H{\mathbb{Z}}\) and consider the long exact sequence in coefficients of \(H^k(X; -)\) with respect to \({\mathbb{Z}}\xrightarrow{\times n} {\mathbb{Z}}\to {\mathbb{Z}}/n\). In the proof of Lemma 5, we showed that \(H^3(X; {\mathbb{Z}}) = {\mathbb{F}}_2\) and \(H^4(X; {\mathbb{Z}}) = 0\). This turns the sequence to: \[\begin{tikzcd} {...} & {\mathbb{F}_2} & {\mathbb{F}_2} & {H^3(X; \mathbb{Z}/n)} & 0 & {...} \arrow[from=1-1, to=1-2] \arrow["{\times n}", from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \end{tikzcd}\] If \(n\) is even, then \(\times n: {\mathbb{F}}_2 \to {\mathbb{F}}_2\) is the zero map, so \(H^3(X; {\mathbb{Z}}/n) = {\mathbb{F}}_2\). If \(n\) is odd, \(\times n: {\mathbb{F}}_2 \to {\mathbb{F}}_2\) is an isomorphism, so \(H^3(X; {\mathbb{Z}}/n) = 0\). ◻

Proof of Theorem [thm::complex95qca95split](4). The spaces \(K(\mathbf{C}({\mathbb{Z}}^n; {\mathbb{F}}_{2^k}))\) for \(n = 0,1\) are covered by the discussions above. By Proposition 51 and since \(\mathbf{Q}({\mathbb{Z}}^0; {\mathbb{F}}_{p^k}) \cong \Omega \mathbf{Q}({\mathbb{Z}}^1; {\mathbb{F}}_{p^k})\), \(\mathbf{Q}({\mathbb{Z}}^0; {\mathbb{F}}_{p^k})\) has only one homotopy group, so its stable \(k\)-invariants are all \(0\). For \({\mathbb{Z}}^1\), \(\mathbf{Q}({\mathbb{Z}}^1; {\mathbb{F}}_{p^k})\) has homotopy groups concentrated at \(\pi_0\) and \(\pi_2\) and hence split, This is because the unstable \(k\)-invariant between them is \(0\) as \(K(\pi_0, 0)\) is discrete. ◻

Proof of Proposition 54. This becomes a special case of Proposition 62 as there is a coarse equivalence \({\mathbb{Z}}^n \cong H \times {\mathbb{Z}}\), and \(H^+\) can be identified as \(H \times {\mathbb{Z}}_{\geq 0}\). ◻

Proof. It suffices to prove each map \(K(\mathbf{C}(X \times \{0\})) \to K(\mathbf{C}(X \times \{0, 1, ..., n\}))\) is an equivalence. Here we use the explicit definition of \(\mathbf{C}(X')\) (Definition 3.1 of [5]). For any space \(X'\) as in Section 2.1, the category \(\mathbf{C}(X')\) has objects \(A(X', q)\) for every quantum spin system \(q: X' \to {\mathbb{Z}}_{>0}\) and morphisms as locality-preserving isomorphisms. \(\mathbf{C}(X')\) is symmetric monoidal under pointwise tensor product. The natural functor \[F_n: \mathbf{C}(X \times \{0\}) \to \mathbf{C}(X \times \{0, 1, ..., n\})\] is then lax symmetric monoidal. By (2.3) of [31], any lax symmetric monoidal functor \(\phi: \mathcal{S}_1 \to \mathcal{S}_2\) between small symmetric monoidal categories that induce an equivalence on \(B\mathcal{S}_1 \to B\mathcal{S}_2\) would induce an equivalence \(K(\mathcal{S}_1) \to K(\mathcal{S}_2)\). We wish to show this for \(F_n\).

We now apply Quillen’s Theorem A to \(F_n\). For any \(D = \mathcal{A}(X \times \{0, 1, ..., n\}, q) \in \mathbf{C}(X \times \{0, 1, ..., n\})\), it suffices to show \(B(d \downarrow F_n)\) is contractible. The category \(D \downarrow F_n\) has objects \(\alpha: D \to F_n(C) \in \mathbf{C}(X \times \{0, 1, ..., n\})\) and morphisms from \(\alpha: D \to F_n(C)\) to \(\alpha': D \to F_n(C')\) is a map \(F_n(f): F_n(C) \to F_n(C')\) such that \(F_n(f) \circ \alpha = \alpha'\).

Now write \(q'(x, k) = 1\) for \(k > 0\) and \(q'(x, 0) = \prod_{j = 0}^n q(x, j)\), and \(D' = \mathcal{A}(X \times \{0, 1, ..., n\}, q')\). Clearly \(D'\) is in the image of \(F_n\). Let \(\alpha: D \to D'\) a locality preserving isomorphism that stacks the tensor factors on \(x \times \{0, 1, ..., n\}\) to \(x \times \{0\}\). We claim \(\alpha\) is an initial object in \(d \downarrow F_n\). Indeed, for any \(\alpha': D \to F_n(C')\), since \(\{0, 1, ..., n\}\) is bounded, one can recreate the same map \(\alpha'\) to in the flattened strip \(X \times \{0\}\), which gives the desired morphism. Uniqueness follows since for any \(F(f), F(g)\) such that \(F(f) \circ \alpha = F(g) \circ \alpha\), \(F(f) = F(g)\) as \(\alpha\) is an isomorphism, and \(f = g\) as the inclusion of the QCA group to a larger set is an injection. Thus, \(B(d \downarrow F_n)\) is contractible, and the proof follows. ◻

Proof of Proposition 62. We check the algebraic case as the unitary case follows similarly. By Lemma 66, we may without loss replace the fiber in the fibration sequence (8 ) into \[\operatorname{hocolim}_n B\mathcal{Q}(X \times \{0, ..., n\})^+ \to B\mathcal{C}(X \times {\mathbb{Z}}_{\geq 0})^+ \xrightarrow{f} B\mathcal{C}(X \times {\mathbb{Z}})^+.\] We first replace \(f\) in the fibration sequence by a strict fibration. This is given by the pullback of \(f: B\mathcal{C}(X \times {\mathbb{Z}}_{\geq 0})^+ \to B\mathcal{C}(X \times {\mathbb{Z}})^+\) and the map from the path space \(ev_0: PB\mathcal{C}(X \times {\mathbb{Z}})^+ \to B\mathcal{C}(X \times {\mathbb{Z}})^+\) that evaluates a path at \(0\). Call the pullback \(E'\), this gives a fibration \[\operatorname{hocolim}_n B\mathcal{Q}(X \times \{0, ..., n\})^+ \to E' \to B\mathcal{C}(X \times {\mathbb{Z}})^+.\] Now consider Construction 64 on this sequence and the plus-construction map \(\psi: B\mathcal{C}(X \times {\mathbb{Z}}) \to B\mathcal{C}(X \times {\mathbb{Z}})^+\), whose criterion is satisfied. By our discussions above, \([\gamma^2]\) is the identity map \(H_2(B\mathcal{C}(X \times {\mathbb{Z}})^+) \to H_2(B\mathcal{C}(X \times {\mathbb{Z}})^+)\). The homology isomorphism given by the plus-construction map ensures \(\psi^*(\gamma^2)\) is the identity map from \(\iota: H_2(B\mathcal{C}(X \times {\mathbb{Z}})) \to H_2(B\mathcal{C}(X \times {\mathbb{Z}}))\). Thus, we see that \(\iota\) is the class representing both universal central extension and primary obstruction.

By Theorem 65, the primary obstruction class is independent of the choice of lift on the 1-skeleton. Thus, it suffices for us to show that \(\bar{\tau}_{\iota}\) is a 2-cocycle that occurs with a particular lift on the 1-skeleton. Indeed, for a map \(S^1 \to B\mathcal{C}(X \times {\mathbb{Z}})^+\), a lift of \(S^1\) to \(E'\) is precisely a homotopy from \(S^1 \to B\mathcal{C}(X \times {\mathbb{Z}})^+\) to a loop \(S^1 \to B\mathcal{C}(X \times {\mathbb{Z}}_{\geq 0})^+ \to B\mathcal{C}(X \times {\mathbb{Z}})^+\).

The map \(\psi: B\mathcal{C}(X \times {\mathbb{Z}}) \to B\mathcal{C}(X \times {\mathbb{Z}})^+\) is the inclusion of a subcomplex, and a 1-cell in \(B\mathcal{C}(X \times {\mathbb{Z}})\) is given as a loop \(\ell_{g}\) representing \(g \in B\mathcal{C}(X \times {\mathbb{Z}})\). For each \(g\), we choose a lift by homotopy from the loop representing \(g \in B\mathcal{C}(X \times {\mathbb{Z}})^+\) to a choice of \(\ell_{t_+(g)} \to B\mathcal{C}(X \times {\mathbb{Z}}_{\geq 0})^+\). This is the lift on the 1-skeleton. A 2-cell in \(B\mathcal{C}(X \times {\mathbb{Z}})\) is given by a triangle whose edges are \(g, h, gh\). Going along the boundary of the triangle gives exactly \(t_+(g) t_+(h) t_+(gh)^{-1}\), where we need to flip the last edge. The element \(t_+(g) t_+(h) t_+(gh)^{-1}\) is supported on \(X \times \{0, ..., m\}\) for some \(m\) sufficiently large and hence naturally restricts to the image of the fiber. Thus, \[\gamma^2(g, h) = t_+(g) t_+(h) t_+(gh)^{-1}.\] This shows that a choice of \(\gamma_2\) could have been \(\bar{\tau}_{\iota}\), so their cohomology classes agree. This concludes the proof. ◻

Proof. Note here \(u_3(\varphi)\) exists as the \(u_2\)-obstruction vanish from \(H_2(\mathcal{C}^*({\mathbb{Z}}^1)) = 0\) (see Column 3 of Table ¿tbl:table95q95z?). This means that \(u_3(\varphi) = \varphi^*(\iota_2)\), where \(\iota_2\) is the identity map in \(\mathrm{Hom}( H_3(\mathcal{C}^*({\mathbb{Z}}^1), H_3(\mathcal{C}^*({\mathbb{Z}}^1))\cong H^3(\mathcal{C}^*({\mathbb{Z}}^1); H_3(\mathcal{C}^*({\mathbb{Z}}^1))\). The class \([\omega(\bullet)]\) is also natural, i.e., for any QCA representations \(\Phi: G' \to \mathcal{C}^*({\mathbb{Z}}^1) \subseteq \mathcal{Q}^*({\mathbb{Z}}^1)\) and map \(f: G'' \to G'\), we have \[[\omega(\Phi \circ f)] = f^*([\omega(\Phi)]).\] Now consider the inclusion map \(i: \mathcal{C}^*({\mathbb{Z}}^1) \to \mathcal{Q}^*({\mathbb{Z}}^1)\). The class \[[\omega(i)] \in H^3(\mathcal{C}^*({\mathbb{Z}}^1); U(1)) \cong \operatorname{Hom}(H_3(\mathcal{C}^*({\mathbb{Z}}^1)), U(1))\] defines a map \(\psi: A \mathrel{\vcenter{:}}= H_3(\mathcal{C}^*({\mathbb{Z}}^1)) \to B \mathrel{\vcenter{:}}= U(1)\). The map \(\varphi: G \to \mathcal{C}^*({\mathbb{Z}}^1)\) gives a natural diagram \[\begin{tikzcd} {H^3(\mathcal{C}^*({\mathbb{Z}}^1); A)} & {H^3(\mathcal{C}^*({\mathbb{Z}}^1); B)} \\ {H^3(G; A)} & {H^3(G; B)} \arrow["{\psi_*}", from=1-1, to=1-2] \arrow["{\varphi^*}"', from=1-1, to=2-1] \arrow["{\varphi^*}", from=1-2, to=2-2] \arrow["{\psi_*}", from=2-1, to=2-2] \end{tikzcd}\] with \(\psi_*(\iota_2) = [\omega(i)]\). Now if \(u_3(\varphi) = 0\), then \(0 = \psi_*(u_3(\varphi)) = \psi^* \circ \varphi^*(\iota_2) = \varphi^* \circ \psi_*(\iota_2) = \varphi^*([\omega[i]]) = [\omega(\varphi)]\). ◻

Proof. The main results of [14], [55] proved that \(\varphi\) is stably linearizable in this case. Thus, \(u_3(\varphi) = 0\) from Theorem [thm::anomalies]. ◻