[proofplan]
We exploit excision and the decomposition of a punctured neighbourhood of $y$ into disjoint neighbourhoods of the preimage points. Using the Hausdorff property, we separate the finitely many preimage points into pairwise disjoint open sets, each contributing a local degree. The sum-of-local-degrees formula then follows from the commutative diagram relating $H_d(S^d)$ to the direct sum of local homology groups $H_d(U_i, U_i \setminus \{x_i\})$.
[/proofplan]
[step:Separate the preimage points into disjoint open sets mapping into a common neighbourhood of $y$]
Since $f^{-1}(y) = \{x_1, \ldots, x_k\}$ is finite and $S^d$ is Hausdorff, we can choose pairwise disjoint open neighbourhoods $U_1, \ldots, U_k$ of $x_1, \ldots, x_k$ respectively. By continuity of $f$, each $f(U_i)$ is a neighbourhood of $y$ in the sense that $f^{-1}(V) \cap U_i \neq \varnothing$ for any neighbourhood $V$ of $y$. By shrinking the $U_i$ if necessary, we may choose a single connected open neighbourhood $V$ of $y$ in $S^d$ such that $f(U_i) \subset V$ for all $i$ and $f^{-1}(V) \subset U_1 \cup \cdots \cup U_k$ (this last condition uses the compactness of $S^d \setminus (U_1 \cup \cdots \cup U_k)$ and the fact that $f^{-1}(y)$ is contained in $U_1 \cup \cdots \cup U_k$).
[guided]
Why can we arrange $f^{-1}(V) \subset U_1 \cup \cdots \cup U_k$? The set $K = S^d \setminus (U_1 \cup \cdots \cup U_k)$ is closed in the compact space $S^d$, hence compact. Since $f^{-1}(y) \cap K = \varnothing$ (all preimage points lie in the $U_i$), we have $y \notin f(K)$, and $f(K)$ is compact (continuous image of compact), hence closed in $S^d$. Therefore $V' = S^d \setminus f(K)$ is an open neighbourhood of $y$ with $f^{-1}(V') \subset U_1 \cup \cdots \cup U_k$. We then take $V = V' \cap V_0$ where $V_0$ is any connected neighbourhood of $y$ contained in $V'$ (which exists since $S^d$ is locally path-connected).
The condition $f(U_i) \subset V$ can be achieved by further shrinking $U_i$ to $U_i \cap f^{-1}(V)$.
[/guided]
[/step]
[step:Identify the local degree via excision]
The local degree of $f$ at $x_i$ is defined as the degree of the map
\begin{align*}
f_*: H_d(U_i, U_i \setminus \{x_i\}) \to H_d(V, V \setminus \{y\}),
\end{align*}
where both relative homology groups are isomorphic to $\mathbb{Z}$ by the [Local Homology of Manifolds](/theorems/2253). By excision, $H_d(S^d, S^d \setminus \{y\}) \cong H_d(V, V \setminus \{y\})$ (excising $S^d \setminus V$, whose closure is contained in $S^d \setminus \{y\}$). Similarly, $H_d(U_i, U_i \setminus \{x_i\}) \cong H_d(S^d, S^d \setminus \{x_i\})$ for each $i$. Both of these local homology groups are isomorphic to $\mathbb{Z}$, and under canonical generators, the induced map $f_*$ is multiplication by $\deg(f)_{x_i}$.
[/step]
[step:Build the commutative diagram relating global and local degrees]
Set $W = U_1 \cup \cdots \cup U_k = f^{-1}(V)$. Since the $U_i$ are pairwise disjoint, excision and additivity give
\begin{align*}
H_d(W, W \setminus f^{-1}(y)) \cong \bigoplus_{i=1}^k H_d(U_i, U_i \setminus \{x_i\}) \cong \mathbb{Z}^k.
\end{align*}
Consider the commutative diagram:
\begin{align*}
\begin{array}{ccc}
H_d(S^d) & \xrightarrow{f_*} & H_d(S^d) \\
\downarrow q & & \downarrow q' \\
H_d(S^d, S^d \setminus f^{-1}(y)) & \xrightarrow{f_*} & H_d(S^d, S^d \setminus \{y\}) \\
\uparrow \cong & & \\
\bigoplus_{i=1}^k H_d(U_i, U_i \setminus \{x_i\}) & &
\end{array}
\end{align*}
The map $q'$ is the quotient map from $H_d(S^d)$ to $H_d(S^d, S^d \setminus \{y\})$. Since $S^d \setminus \{y\}$ is contractible, the long exact sequence of the pair $(S^d, S^d \setminus \{y\})$ gives an isomorphism $q': H_d(S^d) \xrightarrow{\sim} H_d(S^d, S^d \setminus \{y\})$. Similarly, the map $q$ factors through $H_d(S^d, S^d \setminus f^{-1}(y))$, and by excision, $H_d(S^d, S^d \setminus f^{-1}(y)) \cong \bigoplus_i H_d(U_i, U_i \setminus \{x_i\})$.
[guided]
The diagram encodes the relationship between the global degree and the local degrees. Here is the logic:
1. The top row is $f_*: H_d(S^d) \to H_d(S^d)$, which is multiplication by $\deg(f)$.
2. The right vertical map $q': H_d(S^d) \to H_d(S^d, S^d \setminus \{y\})$ is an isomorphism because $S^d \setminus \{y\}$ is contractible (homeomorphic to $\mathbb{R}^d$), so the long exact sequence of the pair gives $0 = H_d(S^d \setminus \{y\}) \to H_d(S^d) \xrightarrow{q'} H_d(S^d, S^d \setminus \{y\}) \xrightarrow{\partial} H_{d-1}(S^d \setminus \{y\}) = 0$.
3. The left vertical map sends the generator of $H_d(S^d)$ into the direct sum $\bigoplus_i H_d(U_i, U_i \setminus \{x_i\})$. Tracing the generator through: it maps to a class whose $i$-th component generates $H_d(U_i, U_i \setminus \{x_i\})$.
4. The bottom horizontal map, restricted to the $i$-th summand, is multiplication by $\deg(f)_{x_i}$.
Therefore, the generator of $H_d(S^d, S^d \setminus \{y\})$ on the right equals the sum of the local contributions.
[/guided]
[/step]
[step:Extract the degree formula by tracing the generator through the diagram]
Let $\mu \in H_d(S^d) \cong \mathbb{Z}$ be a generator. Under $q'$, $\mu$ maps to the generator $\mu_y$ of $H_d(S^d, S^d \setminus \{y\}) \cong \mathbb{Z}$. Under $f_*$, $\mu$ maps to $\deg(f) \cdot \mu$, which then maps under $q'$ to $\deg(f) \cdot \mu_y$.
On the other path through the diagram: $\mu$ maps under the left vertical arrow into $\bigoplus_i H_d(U_i, U_i \setminus \{x_i\})$, with the $i$-th component being the local generator $\mu_{x_i}$. Applying $f_*$ to each summand gives $\deg(f)_{x_i} \cdot \mu_y$ in $H_d(V, V \setminus \{y\}) \cong H_d(S^d, S^d \setminus \{y\})$. The total contribution from all summands is $\left(\sum_{i=1}^k \deg(f)_{x_i}\right) \cdot \mu_y$.
By commutativity of the diagram,
\begin{align*}
\deg(f) \cdot \mu_y = \left(\sum_{i=1}^k \deg(f)_{x_i}\right) \cdot \mu_y.
\end{align*}
Since $\mu_y$ generates $\mathbb{Z}$, we conclude
\begin{align*}
\deg(f) = \sum_{i=1}^k \deg(f)_{x_i}.
\end{align*}
[/step]
