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?
[Math] Show that degree of constant map is zero
algebraic-topologyhomology-cohomology
Related Solutions
Yes, this is true. Here's one way to prove it. For $0\leq j<n$, write $s_j^{n-1}:\Delta^n\to\Delta^{n-1}$ for the map induced by the order-preserving surjection $\{0,\dots,n\}\to\{0,\dots,n-1\}$ that maps both $j$ and $j+1$ to $j$. Say that a singular simplex $\Delta^n\to X$ is degenerate if it factors through $s_j^{n-1}$ for some $j$. Note that the boundary of a degenerate simplex is a linear combination of degenerate simplices: all but possibly two of its faces are degenerate (the two exceptions being the faces corresponding to omitting the vertices $j$ and $j+1$), and those two faces cancel out. So if we write $D_n(X)\subseteq C_n(X)$ for the span of the degenerate simplices, $D_\bullet(X)$ is a subcomplex of $C_\bullet(X)$. Write $N_\bullet(X)=C_\bullet(X)/D_\bullet(X)$.
Now note that given an order-preserving simplicial map $f:X\to Y$, the induced maps $C^\Delta_\bullet(X)\to C^\Delta_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)$ and $C^\Delta_\bullet(X)\to C_\bullet(X)\to C_\bullet(Y)\to N_\bullet(Y)$ are equal, since the $n$-simplices $\sigma$ that you're worried about have the property that $f\circ c_\sigma$ is degenerate. So to show that $f^\Delta_{\bullet_*}=f_{\bullet_*}$, it suffices to show that the map $C_\bullet(Y)\to N_\bullet(Y)$ induces isomorphisms on homology. By the long exact sequence in homology associated to the short exact sequence $0\to D_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)\to 0$, it suffices to show that $D_\bullet(Y)$ has trivial homology.
We can show this by constructing a chain homotopy. Given a degenerate $n$-simplex $\sigma:\Delta^n\to Y$, let $j(\sigma)\in\{0,\dots,n-1\}$ be the least $j$ such that $\sigma$ factors through $s_j^{n-1}$. Now define $H:D_n(Y)\to D_{n+1}(Y)$ by $H(\sigma)=(-1)^{j(\sigma)}\sigma\circ s^n_{j(\sigma)}$. An elementary computation then shows that $H\partial+\partial H:D_n(Y)\to D_n(Y)$ is the identity map. It follows that $D_\bullet(Y)$ has no homology.
To give a sense of the computation $H\partial+\partial H=1$, let's show what happens when you take a $3$-simplex $\sigma$ with $j(\sigma)=1$; the general case works very similarly. Let's write $\sigma=[a,b,b,c]$; all the simplices built from $\sigma$ will be written using similar expressions with the obvious meaning (here $a$, $b$, and $c$ are the vertices of $\sigma$, with $b$ repeated since it factors through $s^2_1$). We then have $H(\sigma)=-[a,b,b,b,c]$, so $$\begin{align}\partial H(\sigma)=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,c]+[a,b,b,c]-[a,b,b,b]\\=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,b].\end{align}$$ On the other hand, $\partial\sigma=[b,b,c]-[a,b,c]+[a,b,c]-[a,b,b]=[b,b,c]-[a,b,b]$. We have $H([b,b,c])=[b,b,b,c]$ and $H([a,b,b])=-[a,b,b,b]$. Thus we get $$H(\partial\sigma)=[b,b,b,c]+[a,b,b,b].$$ When we add together $\partial H(\sigma)$ and $H(\partial\sigma)$, all the terms cancel except $[a,b,b,c]$, which is just $\sigma$.
Let $\alpha$ be an $n$-cell of $X$ and $\beta$ be an $n$-cell of Y. Then $f(\alpha)=\sum_{\beta\in J_n'} y_{\alpha \beta} \beta$. Here $J_n'$ denotes the set of $n$-cells of $Y$. We wish to determine the values of $y_{\alpha \beta}\in \mathbb Z$.
Let $\varphi_\alpha:S^{n-1}\to X^{n-1}$ be the attaching map of $\alpha$. Let $\overline{f}:X^n/X^{n-1}\to Y^n/Y^{n-1}$ be the induced map by $f$ on the quotient.
For every wedge sum $\bigvee_i A_i$ of pointed topological spaces there are canonical retractions $r_j: \bigvee_i A_i \to A_j$. Note that for any CW complex $Z$, $Z^n/Z^{n-1}$ is a wedge of $n$-spheres. Under this identification, we obtain retractions $r_\gamma:Z^n/Z^{n-1}\to S^n$, one for every $n$-cell $\gamma$ of $Z$.
With this notation set up, consider the following commutative diagram:
Proposition: $y_{\alpha \beta}=deg(f_{\alpha \beta})$.
This is theorem 10.13 of Switzer's book "Algebraic topology: Homology and Homotopy", or proposition 3.8 of Lundell and Weingram "The topology of CW complexes". They call $y_{\alpha \beta}$ the "degree with which $\alpha$ is mapped into $\beta$ by $f$", which is a nice name.
Best Answer
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.