I'm reading a proof of the following theorem
Let $X$ and $Y$ be compact connected oriented $n$-manifolds, and let $f:X\to Y$ be continuous. Let $y\in Y$, and assume that $f^{-1}(y)$ is finite. Then $$\deg(f)=\sum_{x\in f^{-1}(y)}\deg_x(f).$$
and I have a question about a small step in the argument. ($X$ and $Y$ being oriented manifolds means we've chosen generators $\theta_X\in H_n(X)\cong\mathbb{Z}$ and $\theta_Y\in H_n(Y)\cong\mathbb{Z}$.)
Proof of Theorem: We see that there is a neighborhood $U_y$ of $y$ in $Y$ such that $$f^{-1}(U_y)=\coprod_{x\in f^{-1}(y)}U_x$$ where each $U_x$ is a neighborhood of $x$. We get an open cover of $X$, consisting of $$\{U_x\mid x\in f^{-1}(y)\}\cup\{X\setminus f^{-1}(y)\}$$ The homology class $\theta_X$ is representable by a cycle $$\xi=\sum_{x\in f^{-1}(y)} \sigma_x+\eta,$$ where $\sigma_x\in C_n(U_x)$ and $\eta\in C_n(X\setminus f^{-1}(y))$. (argument continues)
I feel like I'm forgetting or missing something obvious, but why is this bolded statement true?
Here is a proof which works for topological manifolds. The argument is similar to the one in the proof of Mayer-Vietoris sequence for singular homology. There is nothing special here about the fundamental class, it works for all elements of $H_n(M)$, where $M$ is a compact metrizable space (e.g., a topological manifold) and I use singular homology. The group $G$ of coefficients is also irrelevant, and I will suppress it in my notation.
Let $\theta\in C_n(M)$ be a singular $n$-cycle, $$ \theta=\sum_{i=1}^k a_i f_i, a_i\in G, f_i: \Delta^n\to M. $$ Let ${\mathcal U}$ be an open cover of $M$. Since $M$ is compact and metrizable, we pick an auxiliary metric on $M$ and let $\lambda>0$ be a Lebesgue number of the covering ${\mathcal U}$. By compactness of the $n$-simplex $\Delta^n$, all the continuous maps $f_i: \Delta^n\to M$ are uniformly continuous. Therefore, we can triangulate $\Delta^n$ so that for each facet $\sigma_j$ of the triangulation and each $i$, $$ diam(f_i(\sigma_j))<\lambda. $$ Let $\theta'$ be the singular $n$-cycle obtained by restricting the maps $f_i$ to $n$-simplices $\sigma_j$ above. Clearly, $[\theta]=[\theta']\in H_n(M)$.
By construction, each $f_i(\sigma_j)$ is contained in one of the open sets $U_{ij}\in {\mathcal U}$, which we fix from now on. Now, set $$ \theta_U := \sum_{i,j} a_i f_i|\sigma_j, $$ where the sum is taken over all pairs $(i, j)$ such that $U=U_{ij}$. Hence, $$ \theta'=\sum_{U\in {\mathcal U}} \theta_U, $$ where each $\theta_U$ is in $C_n(U)$.