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This might be a stupid question, but I want to make sure as a beginner of AT. A map between CW-complexes $f: A \rightarrow B$ is defined to be a weak homotopy equivalence if it induces isomorphisms $f_*: \pi_n(A) \rightarrow \pi_n(B)$ for all $n$. But is it true that $\pi_n(A) \cong \pi_n(B)$ for all $n$ implies that such a map $f$ exists? If not, what is the reason? And what is a good counterexample?

Clueless Gorilla
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    You might like this MO question: http://mathoverflow.net/questions/55365/counterexamples-in-algebraic-topology Basically, if you haven't seen a theorem saying something, there's probably a counterexample for it, even if your gut screams "it's true!". – Najib Idrissi Oct 05 '16 at 08:03

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No, there are examples of spaces for which all homotopy groups are isomorphic, but not map induces these isomorphisms for all $n$ simultaneously.

For example, let $A = \mathbb{R}P^2\times S^3$ and let $B = S^2\times \mathbb{R}P^3$.

The universal cover of $A$ and $B$ are both $S^2\times S^3$, which implies all the higher homotopy groups are isomorphic. Further, both fundamental groups are isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

So, why isn't there a map which induces an isomorphisms on all $n$? By Whitehead's theorem, if there was such map, then $A$ and $B$ would be homotopy equivalent. They are not since, for example, the top homology group $H_5$ is trivial for $A$ but non-trivial for $B$.

Jason DeVito - on hiatus
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  • Nice, this is a lot simpler than mine. –  Oct 05 '16 at 01:55
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    More fun example: $A = S^2\times S^3$ and $B$, the unique non-trivial linear $S^3$ bundle over $S^2$. Then $A$ and $B$ have the same homotopy groups (both are $S^1$ quotients of $S^3\times S^3$), homology groups, and cohomology ring structure (from the Gysin sequence and Poincare duality). However, the second Stiefel-Whitney class $w_2 \in H^2(A;\mathbb{Z}/2\mathbb{Z})$ is trivial but is non-trivial for $B$. – Jason DeVito - on hiatus Oct 05 '16 at 01:56
  • @MikeMiller: I actually prefer your example - they are horribly not homotopy equivalent. – Jason DeVito - on hiatus Oct 05 '16 at 01:58
  • I like your second example for multiple reasons, one of which being that it implies that if $H$ is the Hopf bundle, the total spaces of $H \oplus \mathbf 2$ and the trivial rank 4 bundle are not proper homotopy equivalent, even though you can't tell them apart via intersection numbers (like you can with eg the Hopf bundle and the trivial rank 2 bundle). –  Oct 05 '16 at 02:01
  • Funny, I just noted the the link to Whitehead's theorem has exactly this counterexample. Well, there is still the one in the comments – Jason DeVito - on hiatus Oct 05 '16 at 02:10
  • Thanks for the explanations. @MikeMiller can you explain a bit more about Hopf bundle and intersection number? Thank you. – Clueless Gorilla Oct 05 '16 at 02:11
  • @CluelessGorilla It's not really worth going into; it's mostly unrelated to your question. –  Oct 05 '16 at 02:13
  • @MikeMiller Ok sure – Clueless Gorilla Oct 05 '16 at 02:14
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Nah. Take, for instance, $S^3 \times \Bbb{CP}^\infty$ and $S^2$. They have the same homotopy groups (check the long exact sequence of the fibration $S^1 \to S^3 \to S^2$) but are not homotopy equivalent (see the homology).

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Another easy/interesting counter example: Consider $X=S^1\vee S^3$ and its double cover $X_2$, i.e attach two copies of $S^3$ one in north pole and one in south pole of $S^1$. Then $\pi_1(X) \cong \mathbb{Z} \cong \pi_1(X_2)$ and the covering map induces isomorphisms in $\pi_n$ for all $n\geq 2$. But $X$ and $X_2$ are not homotopically equivalent since their Euler Characteristics are different. So there cannot be any induced isomorphism by Whitehead theorem.

Hanno
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Anubhav Mukherjee
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