7
$\begingroup$

Let $f \colon S^n \to S^n$ be a constant map, $n > 0$. I want to show that $\deg f = 0$. I will do it by definition. Let $\sum_k g_k \sigma^n_k$ be a singular chain, $g_k \in \mathbb{Z}$, $\sigma^n_k \colon \Delta^n \to S^n$, where $\Delta^n$ is a standard $n$-simplex. Then the induced map $f_n \colon C_n(S^n) \to C_n(S^n)$ acts by the rule $$ f_n \left( \sum_k g_k \sigma^n_k \right) = \left( \sum_k g_k \right) \sigma^n_0, $$ where $\sigma^n_0 = f \colon \Delta^n \to pt$. Let's notice that $\partial \sigma^n_0 = 0$ if $n$ is odd and $\partial \sigma^n_0 = \sigma^{n-1}_0$ if $n$ is even. Hence if $n$ is even the image of cycles group under the action of $f_n$ is trivial and hence the induced homomorphism $f_n \colon H_n(S^n) \to H_n(S^n)$ in homology groups is trivial. But how to show that if $n$ is odd then the induced homomorphism $f_n$ in homology groups will also be trivial?

$\endgroup$
8
  • $\begingroup$ I think that if $n$ is odd then $\partial\sigma^{n+1}_0=\sigma^n_0$, so $\sigma^n_0$ is a boundary. So it is the other way round. $\endgroup$ Commented Apr 2, 2013 at 21:16
  • $\begingroup$ @StefanH. but there are no $(n+1)$-dimensional nontrivial chains, so all $n$-dimensional boundaries are all trivial, isn't it true? Ah, I speak about simplicial chains, not singular. Singular chains may be nontrivial for larger dimensions than $n$, right? Thank you! $\endgroup$
    – Appliqué
    Commented Apr 2, 2013 at 21:19
  • $\begingroup$ Dont't you want to show that $\sigma_0^n$ is a boundary? Then it is trivial in the homology group. $\endgroup$ Commented Apr 2, 2013 at 21:22
  • $\begingroup$ @StefanH. It is clear that $\sigma^n_0$ is a boundary of $\sigma^{n+1}_0$ because $\partial \sigma^{n+1}_0$ is a sum of $n+1$ copies of $\sigma^n_0$ with alternating signs starting from $+$. The problem arised because I thought about simplicial chains when I was thiking about $n$-dimensional boundaries. Now I solved it. $\endgroup$
    – Appliqué
    Commented Apr 2, 2013 at 21:23
  • 2
    $\begingroup$ You can see this very easily by factoring $f$ through the inclusion $S^n - \{ x \} \hookrightarrow S^n$, where $x$ is any point not in the image of $f$: the homomorphism $f_*$ factors through $H_n(S^n - \{x\})$ but $S^n - \{x\}$ is contractible. $\endgroup$
    – user314
    Commented Apr 3, 2013 at 4:51

1 Answer 1

12
$\begingroup$

You can say a little bit more, the following proposition is also proved in Hatcher's Algebraic Topology

Proposition. Let $f:S^n \to S^n$ be a continuous map, if $f$ is not surjective then deg$(f) = 0$

Proof. Choose a point $x_0$ not in the image of $f$, then you can factor $f$ with the inclusion of $S^n\setminus \{x_0\}$ which is homeomorphic to $\mathbb{R}^n$. Applying the functor $H_n(\text{_})$ gives us the required result.

$\endgroup$
1
  • 1
    $\begingroup$ This is the more "elegant" answer. It also works for any ordinary homology theory, not just singular homology. $\endgroup$ Commented May 11, 2017 at 13:25

You must log in to answer this question.

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