# Proof that two homotopy inverses are homotopic

Let $X$ and $Y$ be topological spaces. A continuous mapping $f : X \to Y$ is said to be a homotopy equivalence if there exists $g : Y \to X$ continuous such that $g\circ f$ is homotopic to $id_{X}$ and $f\circ g$ is homotopic to $id_{Y}$. In such a case, $g$ is called a homotopy inverse of $f$.

Show that if $g_{1}$ and $g_{2}$ are homotopy inverses of a mapping $f$, then $g_{1}$ and $g_{2}$ are homotopic.

I know that two continuous functions are called homotopic if there is a function $F:X \times I \to Y$ with the properties $H(x,0) = f(x)$ and $H(x,1) = g(x)$. The first parameter is our original function, the second one shows the extent of deformation.

How I can use these informations to show that $g_{1}$ and $g_{2}$ are homotopic?

If you know that you can compose homotopies, then $$g_1\simeq g_1\circ (f\circ g_2)= (g_1\circ f)\circ g_2\simeq g_2$$ where $g_1$ and $g_2$ are the two homotopy inverses.
• @cvis It is the whole proof as long as you are allowed to assume, if $h\simeq k$, then $f\circ h\simeq f\circ k$. – Joe Johnson 126 Oct 13 '13 at 13:23
First prove that $h \simeq k$ implies $f \circ h \simeq f \circ k$ and $g \circ h \simeq g \circ k$ for all (composable) continuous maps. This allows us to define the homotopy category. Objects are topological spaces, and a morphism is a homotopy class of continuous maps. The composition is defined by $[f] \circ [g] := [f \circ g]$.
But now we can simply observe that, in any category, the inverse of a morphism is uniquely determined: If $g_1$ and $g_2$ are inverse to $f$, then $$g_1 = g_1 \circ \mathrm{id} = g_1 \circ (f \circ g_2) = (g_1 \circ f) \circ g_2 = \mathrm{id} \circ g_2 = g_2$$ Conclusion: Essentially this has nothing to do with homotopy.