Simultaneous CW Approximation

Question

Given a topological space $X$, we know that there is a CW complex $Z$ with a map $Z\rightarrow X$ inducing an isomorphism on homotopy groups.

If we are given two spaces $X_{1}$ and $X_{2}$ with isomorphic homotopy groups (but no continuous map inducing the isomorphism), it makes sense that there shouldn't be a simultaneous CW approximation of $X_{1}$ and $X_{2}$, that is, there shouldn't be a CW complex $Z$ with maps to both $X_{1}$ and $X_{2}$ inducing isomorphisms on homotopy groups.

So why does the naive approach to use the method in Hatcher (Proposition 4.13) not work? Assume without loss of generality everything is path connected and our spaces are pointed.

1) Let $Z_{0}$ be a point and we fix an inclusion of $Z_{0}$ into $X_{1}$ and $X_{2}$.

2) Now, let $\{\gamma_{\alpha}\}$ be generators of $\pi_{1}(X_{1})$. Let $\{\gamma_{\alpha}'\}$ be the corresponding generators in $\pi_{1}(X_{2})$ under the isomorphism $\pi_{1}(X_{1})\rightarrow \pi_{1}(X_{2})$ (that is not induced by a continuous map).

3) So now, attach a copy of $S^{1}$ to $Z_{0}$ for each generator $\gamma_{\alpha}$ and map $S^{1}$ to $X_{1}$ as a representative of the class of $\gamma_{\alpha}$ in $\pi_{1}(X_{1})$. For the \emph{same} copy of $S^{1}$, map it to the representative of the class of $\gamma_{\alpha}'$ in $\pi_{1}(X_{2})$.

4) Let this new space be $Z_{1}$. By construction, the map $Z_{1}\rightarrow X_{1}$ induces a surjection $\pi_{1}(Z_{1})\rightarrow \pi_{1}(X_{1})$ and similarly, $Z_{1}\rightarrow X_{2}$ induces a surjection $\pi_{1}(Z_{1})\rightarrow \pi_{1}(X_{2})$.

5) For each element $\alpha$ of the kernel of $\pi_{1}(Z_{1})\rightarrow \pi_{1}(X_{1})$ , attach a disk $D^{2}$, where the boundary map is a representative of $\alpha$. From the isomorphism $\pi_{1}(X_{1})\rightarrow\pi_{1}(X_{2})$, $\alpha$ is also in the kernel of $\pi_{1}(Z_{1})\rightarrow \pi_{1}(X_{2})$.

6) Let the new space obtained after attaching these disks be $Z_{1}'$.

7) Now, we want to attach copies of $S^{2}$ so that our maps will induce surjections on $\pi_{2}$. We do the same thing as we did for step three to get a space $Z_{2}$.

8) Continue this process indefinitely to build a $CW$ approximation $Z$ of $X_{1}$ and $X_{2}$ such that $Z\rightarrow X_{1}$ induces isomorphisms $\pi_{*}(Z)\rightarrow \pi_{*}(X_{1})$ and $\pi_{*}(Z)\rightarrow \pi_{*}(X_{2})$.

Answer 1

As you say there is often not even a zigzag of weak equivalences between two arbitrary spaces with the same homotopy groups.

See https://mathoverflow.net/questions/3540/are-there-two-non-homotopy-equivalent-spaces-with-equal-homotopy-groups for some examples. These are examples that are CW complexes, and so if there were a zig-zag of weak equivalences there would actually be a homotopy equivalence (Whitehead's theorem).

I would guess the problem is that you don't have any way of extending the particular map constructed in steps 1-6 to a map between the spaces constructed in step 7.

Answer 2

Your construction looks fine to me, however you have not actually shown that there exists a weak homotopy equivalence between $X_1$ and $X_2$.

Let $\stackrel{\sim}{\rightarrow}$ denote the relation of weak equivalence. You have shown that $Z\stackrel{\sim}{\rightarrow}X_1$ and $Z\stackrel{\sim}{\rightarrow}X_2$ but, because $\stackrel{\sim}{\rightarrow}$ is not a symmetric relation, we can not deduce that $X_1\stackrel{\sim}{\rightarrow}X_2$ without further assumptions (normally CW assumptions).

I'm not sure on the history of the subject, but I imagine maps with these properties are only called equivalences because they really do induce equivalence relations in some categories.