Equivalence class of composition of a loop and a homeomorphism

Question

How do I solve the following: Let $X$ be a topological space with base point $x_0$ and $f$ be a loop at $x_0$. Let $h:[0,1] \rightarrow [0,1]$ be a homeomorphism. Prove that either $[f \circ h] = [f]$ or $[f \circ h] = [f]^{-1}$. I tried showing that there is a path homotopy between $f\circ h$ and $f$ or $f^{-1}$ but that did not work out.

Answer 1

Let $h : [0, 1] \to [0, 1]$ be a homeomorphism. Then either $h(0) = 0$ and $h(1) = 1$, or $h(1) = 0$ and $h(0) = 1$. This is because $0$ and $1$ are the only points $p$ such that $h \setminus \{p\}$ is connected.

I will only handle the case where $h(0) = 0$ and $h(1) = 1$. Then define $\gamma_t(s) = f(t h(s) + (1 - t) s)$. We see that $\gamma : [0, 1]^2 \to X$ is continuous. And we see that $\gamma_0 = f$, $\gamma_1 = f \circ h$. Thus, $[f] = [f \circ h]$.