[proofplan] We prove each of the four properties separately. Property (1) follows from the definition of the induced map on homology. Property (2) uses a factorisation of a non-surjective map through the contractible space $S^n \setminus \{p\}$. Property (3) is immediate from the functoriality of the induced map. Property (4) follows from the [homotopy invariance of homology](/theorems/2237), which guarantees that homotopic maps induce the same homomorphism on each homology group. [/proofplan] [step:Prove that $\deg(\operatorname{id}_{S^n}) = 1$] The induced map $(\operatorname{id}_{S^n})_*: H_n(S^n) \to H_n(S^n)$ is the identity homomorphism, since the identity map on a space induces the identity on all homology groups (this follows directly from the definition: $(\operatorname{id}_{S^n})_n(\sigma) = \operatorname{id}_{S^n} \circ \sigma = \sigma$ for every singular $n$-simplex $\sigma$). The identity homomorphism on $H_n(S^n) \cong \mathbb{Z}$ is multiplication by $1$, so $\deg(\operatorname{id}_{S^n}) = 1$. [/step] [step:Prove that $\deg(f) = 0$ when $f$ is not surjective, by factoring through a contractible space] Suppose $f: S^n \to S^n$ is continuous and there exists $p \in S^n \setminus f(S^n)$. Then $f$ factors as \begin{align*} S^n \xrightarrow{\;f\;} S^n \setminus \{p\} \hookrightarrow S^n, \end{align*} where the first map is $f$ with codomain restricted to $S^n \setminus \{p\}$ (which is valid since $p \notin f(S^n)$), and the second map is the inclusion. By stereographic projection, $S^n \setminus \{p\} \cong \mathbb{R}^n$, which is contractible. Therefore $H_n(S^n \setminus \{p\}) = 0$ for $n \geq 1$. By functoriality of the induced map: \begin{align*} f_*: H_n(S^n) \to H_n(S^n \setminus \{p\}) \to H_n(S^n). \end{align*} The first map lands in $H_n(S^n \setminus \{p\}) = 0$, so the composition $f_*$ is the zero homomorphism. Since $f_*$ is multiplication by $\deg(f)$ and $f_*$ is the zero map, $\deg(f) = 0$. [guided] The idea is that a map $f: S^n \to S^n$ that misses a point $p$ has its image contained in $S^n \setminus \{p\}$, which is homeomorphic to $\mathbb{R}^n$. Since $\mathbb{R}^n$ is contractible, it has trivial $n$-th homology, and any map that factors through a space with $H_n = 0$ must induce the zero map on $H_n$. More precisely, the functoriality of homology gives us the factorisation $f_* = \iota_* \circ \tilde{f}_*$, where $\tilde{f}: S^n \to S^n \setminus \{p\}$ is the codomain restriction and $\iota: S^n \setminus \{p\} \hookrightarrow S^n$ is the inclusion. Since $\tilde{f}_*: H_n(S^n) \to H_n(S^n \setminus \{p\}) = 0$ maps into the trivial group, $\tilde{f}_*$ is the zero map, and hence $f_* = \iota_* \circ 0 = 0$. Why does stereographic projection work here? The map that sends $(x_1, \ldots, x_{n+1}) \in S^n \setminus \{p\}$ to a point in $\mathbb{R}^n$ by projecting from $p$ is a homeomorphism (with explicit inverse). The contractibility of $\mathbb{R}^n$ then gives $H_n(S^n \setminus \{p\}) \cong H_n(\mathbb{R}^n) = 0$. [/guided] [/step] [step:Prove $\deg(f \circ g) = \deg(f) \cdot \deg(g)$ using functoriality] By the [functoriality of induced maps on homology](/theorems/2233), $(f \circ g)_* = f_* \circ g_*$. Both $f_*$ and $g_*$ are endomorphisms of $H_n(S^n) \cong \mathbb{Z}$, so they are multiplication by $\deg(f)$ and $\deg(g)$ respectively. The composition of multiplication by $\deg(f)$ and multiplication by $\deg(g)$ on $\mathbb{Z}$ is multiplication by $\deg(f) \cdot \deg(g)$: \begin{align*} (f \circ g)_*(\alpha) = f_*(g_*(\alpha)) = f_*(\deg(g) \cdot \alpha) = \deg(f) \cdot \deg(g) \cdot \alpha \end{align*} for all $\alpha \in H_n(S^n)$. Therefore $\deg(f \circ g) = \deg(f) \cdot \deg(g)$. [/step] [step:Prove that homotopic maps have equal degree] If $f \simeq g$, then by [Homotopy Equivalences Induce Isomorphisms](/theorems/2237) (or more precisely, the underlying fact that [homotopic maps induce equal maps on homology](/theorems/???)), $f_* = g_*: H_n(S^n) \to H_n(S^n)$. Since both are multiplication by their respective degrees on $H_n(S^n) \cong \mathbb{Z}$, $\deg(f) = \deg(g)$. [guided] This property says that degree is a homotopy invariant. The underlying algebraic fact is that homotopic maps $f \simeq g: X \to Y$ induce the same homomorphism $f_* = g_*: H_n(X) \to H_n(Y)$ for every $n$. The proof of this fact constructs a chain homotopy between the chain maps $f_\#$ and $g_\#$ using a prism operator, so that on the level of cycles, $f_\#(c) - g_\#(c)$ is always a boundary, hence $[f_\#(c)] = [g_\#(c)]$ in homology. Applied to $X = Y = S^n$: $f_*$ and $g_*$ agree as endomorphisms of $H_n(S^n) \cong \mathbb{Z}$. Since an endomorphism of $\mathbb{Z}$ is determined by its value on a generator (it must be multiplication by some integer), and $f_*$ is multiplication by $\deg(f)$ while $g_*$ is multiplication by $\deg(g)$, equality $f_* = g_*$ forces $\deg(f) = \deg(g)$. [/guided] [/step]