Every homeomorphism of the torus is homotopic to a homeomorphism with a fixed point Let $f : T^2 \to T^2$ be a self-homeomorphism of the torus. I'm trying to prove that there exist a homeomorphism $g: T^2 \to T^2$ with a fixed point, such that $f$ and $g$ are homotopic.
My initial attempt was to lift $f$ to a homeomorphism $\tilde{f}$ of the universal cover $\mathbb{R}^2$ of $T^2$, and restrict $\tilde{f}$ to a fundamental domain of the covering $\mathbb{R}^2 \to T^2$ given by the unit square $U$. Since the unit square is homeomorphic to the closed disk, the domain and image of $\tilde{f}|_U$ are homeomorphic to a closed disk. One can then apply the Brouwer fixed point theorem to the composition of $\tilde{f}|_U$ with the homeomorphisms. However, I'm not sure how I can use this to find a map homotopic to $\tilde{f}$ (and such that the homotopy is equivariant under the action of the deck group on $\mathbb{R}^2$) with a fixed point.
 A: You don't need the Brouwer fixed point theorem, you just need a very concrete construction.
The key fact is that for every two points $P,Q \in T^2$ there exists a homeomorphism $g : T^2 \to T^2$ such that $g(P)=Q$ and such that $g$ is homotopic to the identity. Once you know that, choosing any $Q \in T^2$ and letting $P = f(Q)$, it follows that $g \circ f$ is homotopic to $f$ and $g \circ f(Q) = Q$.
The function $g$, and the homotopy, can be defined first up in the universal cover. Choose lifts $\tilde P, \tilde Q \in \mathbb R^2$ of $P,Q$. Now just do a straight line homotopy, sliding each point along the displacement vector $\tilde Q - \tilde P$:
$$\tilde H(a,t) = a + (1-t)(\tilde Q-\tilde P)
$$
and so $\tilde g(\tilde P)=\tilde H(\tilde P,0)=\tilde Q$, and $\tilde H(\tilde P,1)=\tilde P$.
Now you have to show that this straight line homotopy upstairs induces a homotopy downstairs. More precisely, letting $[a]=a+\mathbb Z^2 \in T^2$ denote the orbit of $a$ under the action of the deck group, you have to check that the formula
$$H([a],t) = [a+(1-t)(\tilde Q- \tilde P)]
$$
gives a well-defined homotopy on $T^2$, and that the function $g([a]) = H([a],0)$ satisfies $g(P)=Q$, and $g([a],1)=[a]$ for all $[a] \in T^2$.
