Degree of a smooth proper function, smooth homotopy and their pullbacks

Question

Let $M$ and $N$ be compact manifolds, and $F, G: M \rightarrow N$ two smooth maps such that $F \sim G$, where $\sim$ means smooth homotopic. Then a well-known result is that the degree of $F$ is the same as the degree of $G$.

A corollary is the following: let $F :M \rightarrow M$, then if $F \sim id_M$, where $id_M$ is the identity function on $M$, then the degree of $F$ is 1. My question is if the converse hold as well. Is it true that if we have a map from a compact manifold to itself with degree 1, then this must be homotopic to the identity?

My attempt: if we know that the degree of $F$ is 1, then we have, by definition of degree $$\int_M F^{\star}\omega = \int_M \omega$$ Hence we must have that $F^{\star}$, the pullback of $F$, is equal to the pullback of the identity. Does the fact that two functions have the same pullback imply that they are smooth homotopic? Thanks!

Answer 1

Consider the map $F:S^1\times S^1\rightarrow S^1\times S^1$ which is the product of the degree $-1$ maps $f:S^1\rightarrow S^1$. The fundamental class of $S^1\times S^1$ is the product $\Omega=[S^1]\otimes[S^1]\in H_2(S^1\times S^1)\cong H_1(S^1)\otimes H_1(S^1)\cong\mathbb{Z}$ and we have $$F_*\Omega=f_*[S^1]\otimes f_*[S^1]=(-[S^1])\otimes(-[S^1])=\Omega.$$ Still, $F$ is not homotopic to the identity, since it does not induce the identity on $H_1(S^1\times S^1)\cong\mathbb{Z}\oplus\mathbb{Z}$.

I'll leave it up to you to rewrite the above in terms of forms if that is how you would prefer to see it.