Let $n\ge 2$. How can you prove that for every continuous $f:S^n\to T^n$, the induced map on singular homology $f_\star:H_n(S^n)\to H_n(T^n)$ is the zero map? Here, $S^n$ is the $n$ dimensional sphere, and $T^n=(S^1)^n$ is the $n$ dimensional torus.

I don't even know where to begin... This was on the final for my graduate course in algebraic topology. We've covered the fundamental group, covering spaces, and the basics for homology (simplicial complexes, Mayer-Viatoris, degree theory, etc). A hint in the right direction would be greatly appreciated.

  • $\begingroup$ Is or was on the final? $\endgroup$ – Jessica B Dec 15 '14 at 7:04
  • $\begingroup$ Whoops, I meant was, the final is now over. $\endgroup$ – Mike Earnest Dec 15 '14 at 7:05

Hint: for $n\geq 2$, the $n$-sphere $S^n$ is simply connected, and thus, any map $f:S^n\to T^n$ admits a lift to the universal cover of the torus, that is to say, there is a map $F:S^n\to \Bbb R^n$ such that $f=\pi\circ F$ where $\pi:\Bbb R^n\to T^n$ is the universal cover of the torus.

  • 2
    $\begingroup$ and then $f_\star=\pi_\star\circ F_\star$, but $F_\star=0$ since $H_n(\mathbb R^n)=0$. Thank you! $\endgroup$ – Mike Earnest Dec 15 '14 at 7:12
  • $\begingroup$ @MikeEarnest No problem : ) $\endgroup$ – Olivier Bégassat Dec 15 '14 at 7:15
  • 1
    $\begingroup$ This is an old question, but so far all this tells us is $f_* = 0$. That a priori does not imply that $f$ is null homotopic, does it? (We only have the converse: that $f \simeq g \Longrightarrow f_* = g_*$ for maps $f, g : X \to Y$) $\endgroup$ – D Ford Jul 21 '17 at 22:20

As an alternative, one could argue in the following way:

It holds for $n\ge2$ $$[S^n,T^n]=[S^n,K(\mathbb{Z},1)\times...\times K(\mathbb{Z},1)]\cong[S^n,K(\mathbb{Z},1)]^n\cong H^1(S^n,\mathbb{Z})^n=0,$$ so every map $S^n\rightarrow T^n$ is even nullhomotopic.

  • $\begingroup$ Doesn't Olivier's proof also show that $f$ is null-homotopic? Lift it to the universal cover, write down a null-homotopy there, then project the null-homotopy back down to $T^n$. $\endgroup$ – user98602 Dec 15 '14 at 16:45
  • $\begingroup$ Yes, it does. My version does not give anything new and is just an alternative. $\endgroup$ – archipelago Dec 15 '14 at 17:00
  • $\begingroup$ Thanks - just checking I wasn't missing something. $\endgroup$ – user98602 Dec 15 '14 at 17:00

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.