Let f: $\mathbb{T}^2 \to \mathbb{T}^2$ be a continuous map, homotopic to the identity map $\text{Id}_{\mathbb{T}^2}$. Is it true that $f$ is surjective?
(The argument for $\mathbb{S}^2$, instead of $\mathbb{T}^2$ doesn't work here, since the torus minus a point is not contractible).
Yes. Here is a more general observation. Let $M$ and $N$ be closed connected oriented $n$-manifolds, so that $H_n(M, \mathbb{Z}) \cong H_n(N, \mathbb{Z}) \cong \mathbb{Z}$ canonically. The degree of a map $f : M \to N$ is the integer $d$ such that $f$ acts by multiplication by $d$ on $H_n$. In particular, the identity $\text{id}_M : M \to M$ has degree $1$.
Proof. We prove the contrapositive. Suppose $f : M \to N$ is not surjective. $N$ minus any point has vanishing $H_n$ by Poincare duality (this is true more generally for any connected noncompact $n$-manifold), and so $f$ induces the zero map on $H_n$: that is, $f$ has degree zero. $\Box$
This is a surprisingly powerful result once you have some tools for computing the degree of a map; for example it can be used to prove the fundamental theorem of algebra, as described on MO.