nLab
homotopy inverse

Let $\sim$ be the relation of being homotopic (for example between morphisms in the category Top). Let $f : X \to Y$ and $g : Y \to X$ be two morphisms. We say that $g$ is a left homotopy inverse to $f$ or that $f$ is a right homotopy inverse to $g$ if $g \circ f \sim {id}_{X}$ . A homotopy inverse of $f$ is a map which is simultaneously a left and a right homotopy inverse to $f$ .

$f$ is said to be a homotopy equivalence if it has a left and a right homotopy inverse. In that case we can choose the left and right homotopy inverses of $f$ to be equal. To show this denote by $g_{L}$ the left and by $g_{R}$ the right homotopy inverse of $f$ . Then

g_{L} \sim g_{L} \circ (f \circ g_{R}) = (g_{L} \circ f) \circ g_{R} \sim g_{R} .

Hence

f \circ g_{L} \sim f \circ g_{R} \sim {id}_{X},

therefore $g_{L}$ is not only a left, but also a right, homotopy inverse to $f$ .

This makes sense in any category equipped with an equivalence relation $\sim$ , which is compatible with the composition (and with the equality of morphisms).

Revised on December 9, 2009 03:55:03 by Toby Bartels (173.60.119.197)

nLab homotopy inverse

nLab
homotopy inverse