I'm trying to prove that the classifying space(BG) of a group $G$ is unique in the homotopy category of CW-complexes.
Here I define "classifying space" as a topological space $X/G$ s.t. $X$ is contractible and $G$ acts freely on $X$. Notice that $\Pi_1(BG) \cong G$. Here is my attempt to prove it(after some hints I saw in the internet and from a friend), please tell me your opinion (especially in the Surjectivity part):
Define two functors $F,G$ from the category homotopy category of CW-complexes to sets:
- $F: X \mapsto Mor(X,BG)$.
- $G: X \mapsto Mor(\Pi_1(X),G)$
I claim that those functors are naturally equivalent, and hence by Yoneda embedding the CW-complex $BG$ is unique up to homotopy. The natural isomorphism is:
$$ Mor(X,BG) \rightarrow Mor(\Pi_1(X),G) $$ $$ f \mapsto f^* $$
Need to show that the isomorphism is natural, injective and surjective (up to homotopy of course). I'm skipping the naturality proof.
Injectivity: Let $f,g\in Mor(X,BG)$ s.t. $f^*=g^*$. By the lifting theorem:
- Since $f^*(X) \subseteq g^*(X)$, we have a unique lift $\tilde{f}$ s.t. $g\circ \tilde{f} = f$.
- Since $g^*(X) \subseteq f^*(X)$, we have a unique lift $\tilde{g}$ s.t. $f\circ \tilde{g} = g$.
Composing $\tilde{f}$ and $\tilde{g}$ again, in both directions, gives unique lifts (from the lifting theorem) that must be identity (from uniqueness). Therefore $\tilde{f}$ and $\tilde{g}$ are homeomorphisms, and in particular $f,g$ are homophonically equivalent.
Surjectivity: This is the part that I'm a little bit scared because I'm not comfortable enough with groupids... It looks like a natural proof though. Here we're finally starting to use the structure of CW complex.
First, observe that for each $n\geq 0$, $Mor(\mathbb{S^n},BG) \cong Mor(\pi_1(\mathbb{S^n}),G)$ by our natural morphism:
- For $n=0$, both $Mor(\mathbb{pt},BG)$ and $Mor(\pi_1(\mathbb{pt}),G)$ have only one homotopy class.
- For $n=1$, $Mor(\mathbb{S^1},BG)$ is $\Pi_1(BG) \cong G$ up to homotopy, and $\pi_1(\mathbb{S^1}) = \mathbb{Z}$ so $Mor(\pi_1(\mathbb{S^1}),G)=Mor(\mathbb{Z},G)=G$.
- For $n\geq 2$, $Mor(\mathbb{S^n},BG)=\pi_n(BG)={e}$ and $Mor(\pi_1(\mathbb{S^n}),G)=Mor({e},G)={e}$
We know that $F$ is right exact (in the meaning that it commutes with pushouts). By Van-Kampen theorem for groupids, $\Pi_1$ commutes with pushouts, hence $G$ is right exact too.
Since we may build a CW-complex by pushouts of sphere, I want to claim that from the surjection on each sphere component we're getting the surjection to the whole CW-complex category.
It seems to me a good proof, but I'm not comfortable enough in this level. Is this proof legal? How much did I lie? And are there any other simplier proofs? (I found only possible solution which uses fiber bundles and didn't understand it enough)