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There is an elementary phenomenon in differential topology, that I've never quite understood: It is well known that the mapping degree (Brouwer degree) $\operatorname{deg}(f) = \operatorname{deg}(f;y)$, $y \in N$, of a smooth map $f \colon M^n \to N^n$ ($M$ closed, $N$ connected) does not depend on the regular value $y$ used to compute it. So if $f$ is not surjective, one may simply pick a point $y \in N \backslash f(M)$ to conclude that $\operatorname{deg}(f) = \operatorname{deg}(f;y) = \sum_{x \in f^{-1}(y)} \operatorname{sign}(f_*)_x = 0$. But how do I know that for another value $z \in N$ with $f^{-1}(z) \neq \varnothing$ the sum $\sum_{x \in f^{-1}(z)} \operatorname{sign}(f_*)_x$ will actually add up to 0? Choosing $y \in N \backslash f(M)$ to conclude that $\operatorname{deg}(f) = 0$ has always appeared like a kind of "proof by definition" to me (using the quite unnatural definition that every $y \in N \backslash f(M)$ is a regular value, although $f^{-1}(y) = \varnothing$) and I have never understood, why it is allowed to make this kind of argument.

Can anybody help me understand this?

difftop
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  • The idea is to pick a path between two points, transverse to $f$. The inverse image of this path under $f$ is a 1-manifold, with boundary $f^{-1}(y) \sqcup f^{-1}(x)$. Now, because the signed count of the number of boundary points of a 1-manifold is zero, you see that the signed count of $f^{-1}(y)$ is the same as that of $f^{-1}(x)$. Nowhere in this proof did we actually use that $f^{-1}(y)$ was nonempty! –  Oct 08 '15 at 14:35
  • Thank you, Mike. That's the kind of explanation I was looking for. Why isn't this argument included in Milnor's "Topology from the Differentiable Viewpoint" or in Hirsch's "Differential Topology"? Milnor doesn't even explain transversality. – difftop Oct 08 '15 at 15:18
  • I don't really know any other way to prove it - are you sure it's not in those? I guess one could instead just prove it coincides with something defined homologically. The proof is, for sure, in Guillemin and Pollack - that's where I learned it. –  Oct 08 '15 at 15:20
  • @MikeMiller: This is not the argument in G&P. They give a local constancy + connectedness proof. Without perturbing $f$, you may not be able to join your two points by a path to which $f$ is transverse. – Ted Shifrin Oct 11 '15 at 03:26
  • @TedShifrin: Because $f$ is compact and $x, y$ are regular values the transversality theory implies that you can pick such a path. (Pick any random path and then homotope it rel endpoints to be transverse to $f$.) I am less confident about where it's in G&P. –  Oct 11 '15 at 05:20
  • @MikeMiller: Check your thinking very carefully. What if we're in dimension $1$ and $f$ is not a local diffeo? I've taught G&P about a dozen times, so I'm quite confident you win't find it. Yes, you can deduce it from transversality extension theorems, but this is a sledge hammer where none is needed. Milnor does it with an isotopy theorem. – Ted Shifrin Oct 11 '15 at 06:02
  • @MikeMiller: I missed the most important point. G&P doesn't even talk about transversality to submanifolds with boundary. Does Hirsch? G&P talk about deforming the map, not target submanifolds (with boundary). – Ted Shifrin Oct 11 '15 at 15:23

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This argument can't be found in Guillemin & Pollack, as per the comments above, but the idea "if you have two oriented sets of points $X$ and $Y$, and you want to show that a signed count of the numbers is equal - find an oriented cobordism between them", can. This works because a signed count of the number of boundary points of a compact 1-manifold is zero, so if you have a compact 1-manifold $N$ with $\partial N = X \sqcup \bar Y$ ($\bar Y$ being $Y$ with the opposite orientation), then $\#X = \#Y$, where here we're taking a signed count.

So you have a smooth map $g: M \to Y$, where $M$ and $Y$ are closed, and you want to show that if $x, y$ are regular values of $g$, then $\# g^{-1}(x) = \# g^{-1}(y)$. Let $f: I \to Y$ be an embedded path with $f(0) = x$, $f(1) = y$. Now invoke the corollary to the transversality theory on page 73 of Guillemin and Pollack to homotope $f$ - without changing what it does at the endpoints! - to a map transverse to $g$. Because you can perturb $f$ by an arbitrary small amount, you can choose it to remain an embedding. Then (by a mild modification of the same argument as usual), because $f(I)$ is transverse to $g$, $g^{-1}(f(I))$ is an oriented submanifold of $M$ with boundary, whose boundary is $f^{-1}(x) \sqcup \overline{f^{-1}(y)}.$

I think what we do here - trying whatever we can to find a cobordism between the two sets of points - is fairly intuitive. In particular, if a map is of degree zero, this shows that for any $x,y$ with $f^{-1}(x) = \varnothing$, there is a manifold whose boundary is $f^{-1}(y)$, hence the signed count of points is zero.