I've recently read about a number of different notions of "degree." Reading over Javier Álvarez' excellent answer for the thousandth time finally prompted me to ask this question:
How exactly do the following three notions of "degree" coincide?
(1) Algebraic Topology. Let $f\colon X \to Y$ be a continuous map between compact connected oriented $n$-manifolds.
Wikipedia tells me that $H_n(X) \cong H_n(Y) \cong \mathbb{Z}$, and that a choice of orientations for $X$ and $Y$ amount to choices of generators $[X], [Y]$ for $H_n(X), H_n(Y)$, respectively. We then define $\deg f$ via $$f_*([X]) = (\deg f)[Y].$$
(2) Differential Topology. Let $f\colon X \to Y$ be a smooth map between oriented $n$-manifolds, where $X$ is compact and $Y$ is connected.
Let $y \in Y$ be a regular value of $f$ (which exists by Sard's Theorem), let $D_xf\colon T_xX \to T_yY$ denote the derivative (a.k.a. pushforward), and define $$(\deg f)_y = \sum_{x \in f^{-1}(y)}\text{sgn}(\det D_xf).$$ It can be shown that $(\deg f)_y$ is independent of the choice of $y \in Y$, so we can talk meaningfully about a single quantity $\deg f = (\deg f)_y$.
(3) Riemann Surfaces. Let $f\colon X \to Y$ be a holomorphic map between compact connected Riemann surfaces.
For $x \in X$, we let $\text{mult}_x(f)$ denote the multiplicity of $f$ at $x \in X$. For $y \in Y$, we define $$(\deg f)_y = \sum_{x \in f^{-1}(y)} \text{mult}_x(f).$$ As in (2), it can be shown that $(\deg f)_y$ is independent of the choice of $y \in Y$. (Does this generalize to arbitrary complex manifolds?)
Thoughts: As was mentioned in my topology class last semester (and also on Wikipedia), there is this concept of "local homology" which lets us compute (1) as a sum of "local degrees."
I imagine that in the case of (2), each of these local degrees is, in fact, equal to $\text{sgn}(\det D_xf)$ because $f$ is a local diffeomorphism at each regular point $x$. I also imagine that in the case of (3), each of these local degrees is, in fact, equal to $\text{mult}_x(f)$ because the degree of $\mathbb{S}^n \to \mathbb{S}^n$, $z \mapsto z^k$ is $k$. (Does this also mean that $f$ is not regular at any point where $\text{mult}(f) \geq 2$? This would make sense, but what is the proof?)
This all seems correct in my head, but I would really like more details if possible.