calculusdifferential-topologygeometrymanifolds
Let $M,N$ be manifolds. Suppose that $f_0, f_1:M\stackrel{C^\infty}\to N$ are homotopic, i.e. there exists a continuous mapping $f:M\times[0,1]\to N$ s.t. $f(x,0)=f_0(x)$, $f(x,1)=f_1(x)$.
Then is there any smooth homotopy $F\in C^\infty(M\times[0,1],N)$ connecting $f_0$ to $f_1$? How to construct it?
Hopefully this is helpful. It is an excerpt from Lee's Introduction to Smooth Manifolds.
For (a): The opposite of $\lim_{\|x\|\to\infty}\|f(x)\|=\infty$ is not “$f$ is bounded”. Rather it is: There is a sequence $x_n$ with $\|x_n\|\to\infty$ such that $\{f(x_n):n\}$ is bounded. Now you can get a contradiction to the compactness similarly as you attempted in Trial 2.
Your proof of (b) is the correct approach, but haven't you learnt a more general theorem about the homotopy invariance of the degree?
The approach to (c) is correct, and it is essentially the only possible approach: You need to use the so-called existence property of the degree (which is also sometimes formulated as part of the so-called additivity property); the homotopy invariance alone is not sufficient. One way to prove that $0$ lies in the image of $f$ is to combine it with the homotopy invariance: Assume by contradiction that $0$ does not lie in the image of $f$. What can you say about $\phi_r$ for $r<r_0$, in particular for $r=0$?
Best Answer
Yes. I assume you mean $M$ and $N$ to be manifolds without boundary, though I expect the result is true when they are, too.
This follows from the fact that, given a manifold with (or without) boundary $f: (W, \partial W)$, a manifold without boundary $N$ and a map $(W, \partial W) \to N$ that's smooth on the boundary, $f$ is homotopic to a smooth map by a homotopy that doesn't modify the map on the boundary. This is known as the Whitney approximation theorem; you can find a proof in e.g. Lee's "Introduction to Smooth Manifolds". Simply take $W = M \times I$ and $f = f_t$; then by hypothesis $f|_{\partial W}$ is smooth.
Note that if we take $W = M$, we get that every continuous map is homotopic to a smooth one; so $[M,N]$ and $[M,N]_{\text{smooth}}$ are in bijection. So there's no interesting difference between "continuous" and "smooth" homotopy theory.