How to prove that the twist map is not homotopic to the identity map?

Question

Suppose $X$ is an non-empty set. Suppose we have the identity map $f:X \times \{0,1\} \rightarrow X \times \{0,1\}$, $(x,i) \mapsto (x,i)$ for $i \in \{0,1\}$, and we have the twist map $g:X \times \{0,1\} \rightarrow X \times \{0,1\}$, $(x,i) \mapsto (x,j)$ for $i,j \in \{0,1\}$ and $i\ne j$. I think that they are not homotopic, but how do I show it?

Answer 1

Let $Y=X\times\{0,1\}$. Recall that a homotopy from the identity $f$ to the twist map $g$ is a continuous map $H:[0,1]\times Y\to Y$ satisfying $H(0,\cdot)=f(\cdot)$ and $H(1,\cdot)=g(\cdot)$.

Let $x\in Y$. The map $t\mapsto H(t,x)$ is a continuous path from $f(x)$ to $g(x)$. In particular $f(x)$ and $g(x)$ belong to the same (path) connected component.

But connected components of $Y=X\times\{0,1\}$ are subsets of either $X\times\{0\}$ or $X\times\{1\}$, whereas $f(x)$ and $g(x)$ will not belong to the same of these two sets, hence a contradiction.