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

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!

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 $$H_1(S^1\times S^1)\cong\mathbb{Z}\oplus\mathbb{Z}$$.
• @TheTurtleHermit correct. While there are manifolds like spheres for which every degree $1$ map is homotopic to the identity, in general a manifold can have degree one maps which are not homotopic to the identity. In fact it is not even known if all degree $1$ self-maps of closed orientable manifolds need be homotopy equivalences. Feb 12, 2021 at 23:26