Let $X = S^n$ and $Y = S^n$ and let $f: X \to Y$ be a continuous map. Suppose that for some $y \in Y$, the preimage $f^{-1}(y)$ consists of finitely many points $x_1, \dots , x_m$. Let $U_1, \dots, U_m$ be disjoint neighbourhoods of $x_1, \dots, x_m$, each mapped by $f$ into a neighborhood $V$ of $y$.
The local degree formula states that
$$ \text{deg }(f)=\sum_i \text{deg }(f|x_i).$$
Here, $\text{deg }(f)$ is the integer of multiplication characterising the homomorphism $f_\star : H_n(X) \to H_n(Y)$. [Since $H_n(X) \cong \mathbb Z$ and $H_n(y) \cong \mathbb Z$, the homomorphism $f_\star$ is multiplication by an integer; this integer is $\text{deg }(f)$.] In a similar manner, $\text{deg }(f|x_i)$ is the integer of multiplication characterising the homomorphism $f_\star : H_n (U_i , U_i – x_i) \to H_n(V, V-y)$.
At first glance, this local degree formula looks like a great computational tool. But when you sit down to apply it to any concrete example, you realise that determining the signs of the $\text{deg }(f|x_i)$'s is a real pain.
First of all, how do we fix the signs of the various degrees appearing in the formula? After all, the signs of the various degrees are only fixed once we fix choices of generators of $H_n(X)$, $H_n(Y)$, $H_n (U_i , U_i – x_i)$ and $H_n (V, V-y)$.
In Hatcher, this issue is totally glossed over. However, after carefully reading the proof of the local degree formula, it is apparent that the signs work out provided that the generators we choose for the $H_n(U_i, U_i – x_i)$ are sign-compatible with the generator we choose for $H_n(X)$, and the generator we choose for $H_n(V, V-y)$ is sign-compatible with the generator we choose for $H_n(Y)$.
The word "sign-compatible" is non-standard terminology that I have invented, but the concept is lifted straight out of the details of Hatcher's proof. Here is its definition:
Let $i_i$ and $j$ denote the natural inclusions of pairs,
$$i_i : (U_i, U_i-x_i) \to (X, X – x_i),$$
$$j: (X, \emptyset) \to (X, X – x_i),$$
and let $(i_i)_\star$ and $j_\star$ denote the homomorphisms they induce on homology,
$$(i_i)_\star : H_n(U_i, U_i – x_i) \to H_n(X , X – x_i),$$
$$j_\star : H_n(X) \to H_n(X, X – x_i).$$
$(i_i)_\star$ and $j_\star$ are in fact isomorphisms: $(i_i)_\star$ is an isomorphism by excision, and $(j_\star)$ is an isomorphism by the long exact sequence for the pair $(X, X – x_i)$.Let $e_i$ be a generator of $H_n(U_i, U_i – x_i)$, and let $e$ be a generator of $H_n(X)$.
We say that $e_i$ and $e$ are sign-compatible if $$(i_i)_\star(e_i) = j_\star(e).$$A similar notion of sign-compatibility applies to generators of $H_n(V, V-y)$ and $H_n(Y)$.
The local degree formula is valid, provided that the various degrees are computed relative to a sign-compatible set of generators of the various homology groups.
In many applications, $f$ maps $U_i$ homeomorphically to $V$, so each $\text{deg }(f |x_i)$ is either $+1$ or $-1$. However, determining the sign requires us to wrangle with the complicated definition of "sign-compatible" generators.
I'd be grateful if somebody could provide some techniques for determining these signs that apply to a broad range of situations.
A concrete example
For illustration, I'll use the local degree formula to compute the degree of the map $f: X \to Y$ with $X = S^1$ and $Y = S^1$ defined by $$f(\cos \phi, \sin \phi) = (\cos 2\phi, \sin 2\phi).$$
This is an example where I can successfully determine the local degrees up to signs, but only by exploiting the high level of symmetry in the setup. My concern is that the technique I've come up with is not generally applicable – it only works in this particular example because of the high level of symmetry.
So let's get into it. Take $y = (1, 0)$. Then $f^{-1}(y) = \{ x_1, x_2 \}$, where $x_1 = (1,0)$ and $x_2 = (-1, 0)$. Take $V$ to be a small open arc around $y$, and take $U_1$ and $U_2$ to be the two connected components of $f^{-1}(V)$ containing $x_1$ and $x_2$ respectively.
$f$ maps each $U_i$ homeomorphically onto $V$, so the two local degrees are either $+1$ or $-1$. We must determine whether they have the same sign (in which case $\text{deg }(f) = \pm 2$), or whether they have opposite signs (in which case $\text{deg }(f) = 0$).
To determine the signs, let $r : X \to X$ be a $180$-degree rotation. We have the commutative diagram
$\require{AMScd}$
\begin{CD}
H_1(X) @>{j_\star}>> H_1(X,X-x_1) @<{(i_1)_\star}<< H_1(U_1, U_1 – x_1)\\
@V{r_\star}VV @V{r_\star}VV @V{r_\star}VV \\
H_1(X) @>{j_\star}>> H_1(X,X-x_2) @<{(i_2)_\star}<< H_1(U_2, U_2 – x_2)\end{CD}
If $e_{U_1}$ is a generator of $H_1(U_1, U_1 – x_1)$ that is "sign-compatible" with a generator $e_X$ of $H_1(X)$, then by our commutative diagram, $e_{U_2} := r_\star(e_{U_1}) \in H_1(U_2, U_2 – x_2)$ is a generator that is "sign-compatible" with the generator $r_\star(e_X) \in H_1(X)$.
But $r : X \to X$, being a rotation, is homotopic to the identity map $X \to X$. So $r_\star(e_X) = e_X$. Thus, the generator $e_{U_2} \in H_1(U_2, U_2 – x_2)$ is actually "sign-compatible" with our original generator $e_X \in H_1(X)$ too.
Now consider the following commutative diagram.
$\require{AMScd}$
\begin{CD}
H_1(U_1, U_1 – x_1) @>{f_\star}>> H_1(V, V-y) \\
@V{r_\star}VV @V{(\text{id})_\star}VV \\
H_1(U_2, U_2 – x_2) @>{f_\star}>> H_1(V, V-y) \end{CD}
By this commutative diagram, we have
$f_\star(e_{U_1}) = f_\star(r_\star(e_{U_1})) = f_\star(e_{U_2}).$
Now let's choose a generator $e_V$ of $H_n(V, V – y)$ and a generator $e_Y$ of $H_n(Y)$ that are "sign-compatible" with one another.
We have $f_\star(e_{U_1}) = \eta e_V$, where $\eta$ is either $+1$ or $-1$. But then, $f_\star(e_{U_2}) = f_\star(e_{U_1}) = \eta e_V$ too.
Therefore, if we define all of our degrees relative to the "sign-compatible" generators $e_{U_1}, e_{U_2}, e_X, e_V, e_Y$, then
$$\text{deg }(f | x_1) = \text{deg }(f | x_2) = \eta,$$
and so, the local degree formula tells us that
$$\text{deg }(f) = 2\eta.$$
In other words, $\text{deg }(f) = \pm 2$.
It took me a bit of time to come up with this technique, and it is somewhat disheartening that my technique only works in a restricted class of scenarios where there is a large amount of symmetry. Therefore, I'd be grateful to learn about other techniques for computing local degrees that are applicable to situations where my technique can't be applied. Of course, I'd also be happy to learn of ways to simplify my argument for the map $f(\cos \phi, \sin\phi) = (\cos 2\phi, \sin 2\phi)$.
Edit: On rereading my calculation for the map $f(\cos \phi, \sin\phi) = (\cos 2\phi, \sin 2\phi)$, I realise that it is inaccurate to claim that the symmetry of the setup is the ingredient that makes my technique work.
Here's a better summary of what's important.
Suppose that $f^{-1}(y)$ consists of two points, $x_1$ and $x_2$. Let $U_1$ and $U_2$ be disjoint neighbourhoods of $x_1$ and $x_2$ respectively, and let $V$ be a neighbourhood of $y$ containing the images of $U_1$ and $U_2$ under $f$.
Suppose that $f|_{U_1} : U_1 \to V$ and $f|_{U_2} : U_2 \to V$ are in fact homeomorphisms. (So in particular, $\text{deg}(f|x_1)$ and $\text{deg}(f|x_2)$ are both $\pm 1$.)
Let $g : U_1 \to U_2$ be the homeomorphism defined as the composition $g = (f|_{U_2})^{-1} \circ f|_{U_1}$.
Suppose, furthermore, that there exists a homeomorphism $\tilde g : X \to X$ extending $g$.
Then, assuming that our degrees are computed relative to a "sign-compatible" set of generators, we have:
- $\text{deg}(f|x_1) = +\text{deg}(f|x_2)$ if $\text{deg}(\tilde g) = +1$.
- $\text{deg}(f|x_1) = -\text{deg}(f|x_2)$ if $\text{deg}(\tilde g) = -1$.
In my example, it is possible to choose $\tilde g$ to be a $180$-degree rotation (the map that I called $r$), and $\tilde g$ happens to be a homeomorphism of the circle to itself with degree $+1$.
The upshot is that the technique I used for $f(\cos \phi, \sin \phi) = (\cos 2\phi, \sin 2\phi)$ is more general than I initially thought – but by no means is it fully general.
Best Answer
Here's an answer that works in the smooth case. That is, we consider $X$ and $Y$ as smooth manifolds, and we assume $f:X\rightarrow Y$ is smooth.
Choose orientations on $X$ and $Y$. For topological manifolds, this means something in homology. However, because things are smooth, we have a few more ways to define an orientation. Here is one.
Fix, once and for all, a volume form $\omega_X$ on $X$ and $\omega_Y$ on $Y$. Being volume forms simply means that $\omega_X$ is a non-vanishing differential $n$-form on $X$, and similary for $Y$.
Then, given a smooth map $f:X\rightarrow Y$, we select a point $y\in Y$ which is a regular value of $f$. This means that for any $x\in f^{-1}(y)$, that the differential $d_x f:T_x X\rightarrow T_y Y$ is surjective. Sard's theorem guarantees that the set of non-regular $y\in Y$ has measure $0$, so finding such a $y$ is typically not difficult.
Since $y$ is a regular value, the pull back $f^\ast \omega_Y$ is injective. In particular, at any $x\in f^{-1}(y)$, the (algebraic) form $(f^\ast \omega_Y)_x$ is non-zero, so is either a positive multiple of $(\omega_X)_x$ or a negative multiple of $(\omega_X)_x$. Positive is the same as local degree $1$, and negative is the same as local degree $-1$.
Before doing an example, I wanted to mention that, in some sense, the above answer is fully general. To see this, first observe that $X$ and $Y$ are naturally smooth manifolds, so that's not really an extra assumption. And while there are continuous non-smooth $f:X\rightarrow Y$, the $C^\infty$ functions are dense in the space of all continuous functions, so any such $f$ is "close" to a smooth one. Since homotopic maps have the same degree, one can, at least in theory, use the above approach on everything.
Now, let me do your example in this language. To begin with, we need to pick $\omega_X$ and $\omega_Y$.
Thinking of $S^1\subseteq \mathbb{R}^2$, I'll let $\omega_X = \omega_Y$ be the restriction of the form $-y dx + x dy$ to $S^1$. Let's quickly verify that this is a volume form (that is, it's never zero).
Fix the point $(x,y)\in S^1$. Then the tangent vector $-y \frac{\partial}{\partial x} + x\frac{\partial}{\partial y}\in T_{(a,b)} S^1$ and $$(-y dx + xdy)\left(-y \frac{\partial}{\partial x} + x\frac{\partial}{\partial y}\right) = y^2 + x^2 = 1\neq 0.$$
Now, consider the point $y = (1,0)$. Let's verify that this is a regular point of $f$. This means that we need to check that for any $x\in f^{-1}(y)$, that $d_x f:T_x S^1\rightarrow T_y S^1$ is surjective. Since the domain and co-domain are both $1$-dimensional real vector spaces, we simply need to show that $d_x f$ is non-zero at any $x\in f^{-1}(y).$
As you noted, this has two preimages, $x_{\pm 1} = (\pm 1,0)$. Consider the two curves $\gamma_{\pm}(t) = \pm(\cos(t),\sin(t))$. Then $\gamma_{\pm}(0) = x_\pm$ and $\gamma_{\pm}'(0) =\pm \frac{\partial}{\partial y}\neq 0$, so we need to verify that $d_{x_{\pm}}f (\gamma_{\pm}'(0))\neq 0$.
This is computed as follows: \begin{align*} d_{x_{+}}f(\gamma_{+}'(0)) &= \frac{d}{dt}|_{t=0} f(\gamma_{+}(t))\\ &= \frac{d}{dt}|_{t=0} (\cos(2t), \sin(2t))\\ &= (-2\sin(2t), 2\cos(2t))|_{t=0}\\ &= 2 \frac{\partial}{\partial y}\\ &\neq 0\end{align*} and
\begin{align*} d_{x_-} f(\gamma_-'(0)) &= \frac{d}{dt}|_{t=0} f(-(\cos(t),\sin(t)))\\ &= \frac{d}{dt}|_{t=0} f((\cos(t+\pi),\sin(t+\pi)))\\ &= \frac{d}{dt}|_{t=0} (\cos(2t + 2\pi), \sin(2t+2\pi))\\ &= (-2\sin(2t+2\pi), 2\cos(2t+2\pi))|_{t=0}\\ &= 2\frac{\partial}{\partial y}\\ &\neq 0.\end{align*}
Thus, $y = (1,0)$ is a regular value.
Finally, we need to compute both $f^\ast((\omega_Y))(\gamma_{\pm}'(0))$ and $\omega_X(\gamma_{\pm}'(0))$ and compare the sign of the answers.
Since $\omega_X = -ydx + xdy$, at the point $x_\pm = \pm(1,0)$, this becomes $(\omega_X)_{x_\pm} = \pm dy$, so $$\omega_X(\gamma_\pm'(0)) = \pm dy\left(\pm \frac{\partial}{\partial y}\right) = 1.$$
On the other hand, $$(f^\ast(\omega_Y))(\gamma_{\pm}'(0)) = \omega_Y( d_{x_{\pm}} f \gamma_{\pm}'(0)) = dy\left( 2 \frac{\partial}{\partial y} \right) = 2.$$
Since $1$ and $2$ have the same sign, the local degrees are both $+1$, so the total degree is $2$.