I have been reading some differential geometry notes and I came a cross the statement that:

if $f:M\rightarrow N$ is homotopic to a constant map, then $f^{*}:H^{k}N\rightarrow H^{k}M$ is the zero map for $k\geqslant 1$

My question is how do I see that $f^{*}$ is indeed the zero map, given that $f$ is homotopic to a constant map.

  • $\begingroup$ Do you know why $f^*$ is the zero map if $f$ is constant? $\endgroup$ Commented May 8, 2017 at 15:15

1 Answer 1


Suppose $g:M\to N$ is a constant map, $g(x)=y\in N$, with $f\sim g$. By the homotopy invariance of cohomology, $f^*:H^k(N;R)\to H^k(M;R)$ and $g^*:H^k(N;R)\to H^k(M;R)$ are the same (for any ring $R$). Since $g=ih$ where $h:M\to\{y\}$ and $i:\{y\}\to N$, it follows that $$ g^*=(ih)^*=h^*i^*=0 $$ since $H^k(\{y\};R)=0$ when $k\geq 1$. Therefore $f^*$ is the zero map.

  • $\begingroup$ Hello Michael! Thanks so much for your reply. Please excuse my ignorance, but why does $H^{k}\left ( \left \{ y \right \}; R \right )=0$ imply that $h^{*}i^{*}=0$? I know it means that every closed form is exact, but why does it imply that $h^{*}i^{*}=0$ ? $\endgroup$ Commented May 10, 2017 at 9:21
  • $\begingroup$ Since $i^*$ is a map from $H^k(N;R)$ to $H^k(\{y\};R)$ (which is $0$), this means $i^*=0$. So then $h^*i^*=0$. $\endgroup$
    – Michael M
    Commented May 10, 2017 at 13:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .