(Munkres) The homotopy relation is an equivalence relation proof: Proving $G(x,t)=F(x,1-t)$ is a homotopy between $f'$ and $f$

Question

In Munkres' Topology, he has the following lemma: The relation $\simeq$ is an equivalence relation. (Where the relation $f \simeq f'$ means $f$ is homotopic to $f'$.)

The part of the proof I am confused about is proving $f \simeq f'$ implies $f' \simeq f$. He says let $F$ be a homotopy between $f$ and $f'$. Then $G(x,t)=F(x,1-t)$ is a homotopy between $f'$ and $f$.

I understand that you sort of start "backwards in time" if we think of $t$ as time, but how do you actually show rigurously that $G(x,t)=F(x,1-t)$ is a homotopy between $f'$ and $f$?

Answer 1

What is a (free) homotopy between $f$ and $g$? It is a continuous function $H:I\times X\to Y$ such that $H(0,x)=f(x)$ and $H(1,x)=g(x)$. Note: I'm using prefix notation $H(t,x)$ which is more popular as far as I'm aware. The order of $I,X$ is irrelevant.

So let $F:I\times X\to Y$ be a homotopy between $f$ and $g$. We want to show that $G(t, x)=F(1-t,x)$ is a homotopy from $g$ to $f$. So lets check properties:

1. Is $G$ continuous? We can write $G$ as the following composition $$I\times X \overset{j\times id}\longrightarrow I\times X\overset{F}\longrightarrow Y$$ where $j(t)=1-t$ is continuous, $id(x)=x$ is the identity (so continuous) and $j\times id$ is the Cartesian product of functions, i.e. $j\times id(x,y)=(j(x),id(y))$. The Cartesian product of continuous functions is always continuous. This concludes that $G$ is continuous.
2. Now we have to check boundary conditions: $$G(0,x)=F(1,x)=g(x)$$ $$G(1,x)=F(0,x)=f(x)$$

Therefore $G$ is a free homotopy from $g$ to $f$.

Additionally if $F$ is a homotopy with basepoint, i.e. $F(t,x_0)=y_0$ for a fixed $x_0, y_0$ and any $t$ then $G$ also preserves the basepoint: $G(t,x_0)=F(1-t,x_0)=y_0$.