There is a question in my homework on the algebraic topology course asking if two spaces $X$ and $Y$ are weakly homotopy equivalent in case for every cellular space $Z$ sets $[Z,X]$ and $[Z,Y]$ are naturally bijective.

I'm wondering if converse statement holds at least for cell complexes. Suppose $f\colon K\to K'$ is weak homotopy equivalence of cell complexes $K,K'$, namely, $f_*\colon \pi_i(K)\to \pi_i(K')$ is isomorphism for all $i$. I need to show that $f_*\colon [X,K]\to [X,K']$ is bijection.

First I tried to show that it is injective. Suppose $f_*[\alpha]=f_*[\beta]$ for some maps $\alpha,\beta\colon X\to K$. I want to prove that $\alpha\simeq \beta$ then. As soon as $f\circ\alpha\simeq f\circ\beta$, then for every spheroid $\varphi\colon S^k\to X$ $f\circ(\alpha\circ\varphi)\simeq f\circ(\beta \circ\varphi)$, and thus $\alpha\circ\varphi\simeq \beta \circ\varphi$, because $f_*$ is bijection between $\pi_k(K)=[S^k,K]$ and $\pi_k(K')=[S^k,K']$.

Suppose $\varphi\colon S^k\to X$ is attaching map for the cell $e^k$ in $X$. I have homotopy $H:S^k\times [0,1]\to K$, $H|_{t=0}=\alpha\circ\varphi$, $H|_{t=1}=\beta\circ\varphi$ I wish I could extend on the whole cell. Indeed, I can use HEP in this case for the homotopy $H$ and map $\alpha|_{D^k}\colon D^k\to K$ and hence I have homotopy $\tilde H\colon e^k\times [0,1]\to K$ between $\alpha$ and $\beta$.

However, there is a problem, because these homotopies are not necessarily agreed. How do I deal with that?

  • $\begingroup$ Oh my gosh, I've just realized that, in fact, all this time I've been trying to prove Whitehead's theorem $\endgroup$ – igortsts Dec 12 '18 at 16:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.