[proofplan]
We verify each of the four properties separately by examining the induced map $f_*: H_n(S^n) \to H_n(S^n)$. Since $H_n(S^n) \cong \mathbb{Z}$, every endomorphism is multiplication by an integer, and the degree is that integer. Property (1) follows from functoriality of the identity. Property (2) uses a factorisation through a contractible space. Property (3) follows from functoriality of composition. Property (4) follows from homotopy invariance of singular homology.
[/proofplan]
[step:Show that the identity has degree $1$]
The identity map $\operatorname{id}_{S^n}: S^n \to S^n$ induces the identity homomorphism $(\operatorname{id}_{S^n})_* = \operatorname{id}_{H_n(S^n)}$ on homology, by functoriality of singular homology. Since $H_n(S^n) \cong \mathbb{Z}$, the identity homomorphism is multiplication by $1$. Therefore $\deg(\operatorname{id}_{S^n}) = 1$.
[/step]
[step:Show that a non-surjective map has degree $0$]
Suppose $f: S^n \to S^n$ is not surjective. Choose a point $p \in S^n \setminus f(S^n)$. Then $f$ factors as
\begin{align*}
S^n \xrightarrow{\tilde{f}} S^n \setminus \{p\} \hookrightarrow S^n,
\end{align*}
where $\tilde{f}$ is $f$ viewed as a map into $S^n \setminus \{p\}$. The space $S^n \setminus \{p\}$ is homeomorphic to $\mathbb{R}^n$ via stereographic projection from $p$, and is therefore contractible. In particular, $H_n(S^n \setminus \{p\}) = H_n(\mathbb{R}^n) = 0$ for $n \geq 1$. By functoriality, the induced map $f_*: H_n(S^n) \to H_n(S^n)$ factors as
\begin{align*}
H_n(S^n) \xrightarrow{\tilde{f}_*} H_n(S^n \setminus \{p\}) = 0 \to H_n(S^n),
\end{align*}
so $f_* = 0$, which means $\deg(f) = 0$.
[guided]
Why does non-surjectivity force the degree to vanish? The key idea is that if $f$ misses even a single point $p$, then $f$ factors through the punctured sphere $S^n \setminus \{p\}$. This punctured sphere is homeomorphic to $\mathbb{R}^n$ (via stereographic projection from $p$), and $\mathbb{R}^n$ is contractible, so all its higher homology groups vanish.
Concretely: since $p \notin f(S^n)$, we can define $\tilde{f}: S^n \to S^n \setminus \{p\}$ by $\tilde{f}(x) = f(x)$, which is well-defined precisely because $f$ never hits $p$. The original map $f$ equals the composition $\iota \circ \tilde{f}$, where $\iota: S^n \setminus \{p\} \hookrightarrow S^n$ is the inclusion.
Applying the homology functor, we get $f_* = \iota_* \circ \tilde{f}_*$. But $\tilde{f}_*: H_n(S^n) \to H_n(S^n \setminus \{p\})$, and the target group is $H_n(\mathbb{R}^n) = 0$ (since $\mathbb{R}^n$ is contractible). So $\tilde{f}_*$ is the zero map, forcing $f_* = 0$, i.e., $\deg(f) = 0$.
[/guided]
[/step]
[step:Prove multiplicativity of degree under composition]
Let $f, g: S^n \to S^n$ be continuous maps. By functoriality of singular homology, $(f \circ g)_* = f_* \circ g_*$. Since $H_n(S^n) \cong \mathbb{Z}$, the map $f_*$ is multiplication by $\deg(f)$ and $g_*$ is multiplication by $\deg(g)$. Their composition $f_* \circ g_*$ is multiplication by $\deg(f) \cdot \deg(g)$. Therefore
\begin{align*}
\deg(f \circ g) = \deg(f) \cdot \deg(g).
\end{align*}
[/step]
[step:Show that homotopic maps have equal degree]
Suppose $f \simeq g: S^n \to S^n$. By homotopy invariance of singular homology, homotopic maps induce equal maps on homology: $f_* = g_*: H_n(S^n) \to H_n(S^n)$. Since both are multiplication by $\deg(f)$ and $\deg(g)$ respectively, we conclude $\deg(f) = \deg(g)$.
[/step]
