Suppose that $e \sim f$ but $f \nsim g$ (i.e. $f$ is not homotopic to $g$). Is it possible that $g \sim e$?

Question

Suppose that $e \sim f$ but $f \nsim g$ (i.e. $f$ is not homotopic to $g$). Is it possible that $g \sim e$?

By $x \sim y$ I mean "$x$ is homotopic to $y$."

Here is my answer, I just want to make sure I'm not missing anything in my logic:

This is not possible. Assume that $g$ were homotopic to $e$. Then since $e \sim f$ by transitivity we have $g \sim e \sim f$ which is a contradiction since $g \nsim f$. Therefore $g \nsim e$.

Is this correct for any functions $e,f,$ and $g$?

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

Your argument is correct. Homotopy of maps is an equivalence relation.