Let $\varphi : M^n \to N^n$ be a proper smooth map between two connected smooth manifolds. Then $\varphi$ induces a linear map $\varphi^* : H_c^n(N) \simeq \mathbb{R} \to H_c^n(M) \simeq \mathbb{R}$ (where $H_c^n$ is the $n$-th de Rham cohomology group with compact support), so there exists $d \in \mathbb{R}$ so that $\varphi^* : x \mapsto d \cdot x$; $d$ is called the degree of $\varphi$. Moreover, it can be shown that if $\varphi : M \to N$ and $\phi : M\to N$ are smoothly homotopic, then $\varphi^*= \phi^*$ hence $\deg(\varphi)= \deg(\phi)$.

However, every proper smooth map $f : \mathbb{R}^n \to \mathbb{R}^n$ and $g : \mathbb{R}^n \to \mathbb{R}^n$ are smoothly homotopic (thanks to $H(t,x)= tf(x)+(1-t)g(x)$), so every proper smooth map $\mathbb{R}^n \to \mathbb{R}^n$ should have the same degree...

It is probably a silly question, but where does my argument fail?

Edit: Sean Eberhard found a first problem; in fact, to conclude that $\varphi^*=\phi^*$ it is needed the homotopy be proper at any $t$. However, here is another contradiction:

If $P(z)=z^n$ (where $\mathbb{R}^2$ and $\mathbb{C}$ are not distinguised), $\deg(P)=n$ and $P$ is proper if $n \geq 1$. Taking $H(t,z)=tz^m+(1-t)z^n$ with $m>n>1$, $m= \deg(H(1, \cdot))= \deg(H(0,\cdot))=n$, a contradiction.

  • $\begingroup$ In $\mathbb{R}^{2n+1}$, $\deg(\operatorname{Id})=1$ but $\deg(-\operatorname{Id})=-1$... $\endgroup$ – Seirios Oct 3 '13 at 10:24
  • $\begingroup$ Does the smooth homotopy have to pass through smooth proper maps? Note that in your homotopy from $\text{Id}$ to $-\text{Id}$, the homotopy passes through the $0$ map. $\endgroup$ – Sean Eberhard Oct 3 '13 at 10:46
  • $\begingroup$ @SeanEberhard: You are right, I edited my question. $\endgroup$ – Seirios Oct 3 '13 at 11:34
  • $\begingroup$ I expect there is still a properness problem. Since we expect $H(t,z)$ to have degree $m$ for $0<t\leq 1$, the problem will be near $t=0$. And indeed $H(t,z)$ has a root $(1/t-1)^{1/(m-n)}$, which blows up as $t\to 0$. So the problem seems to be that $H(t,z)$ is not "proper in the neighbourhood of $t=0$". Maybe you could give a reference for the theorem you are using about smoothly homotopic maps inducing the same map on cohomology. $\endgroup$ – Sean Eberhard Oct 3 '13 at 11:42
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    $\begingroup$ I agree with Sean: the standard degree theory (including the homotopy invariance) only works in the compact setting. If you generalize to the cohomology with compact support you have to require proper maps everywhere. The proof of the homotopy invariance goes through the pullback $H^*$ of $\omega$ to $I \times M$ and without properness the required integral in the proof is simply undefined. $\endgroup$ – Marek Oct 3 '13 at 12:44

In the interest of marking this question as answered, I'll just say that properness is important throughout the homotopy. See the comments for discussion.


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