---
tag: 2E24G3S4CK
kind: definition
name: Homotopy
msc: [55Q05, 55P10]
defines: 9YA6S6HD9V
uses: [0CQHFPGJ45, 0VGFHTM75Y, 0GBG8B49AA]
sources:
  - url: https://en.wikipedia.org/wiki/Homotopy
---

## Definition

Let $X$ and $Y$ be [[topological spaces|0CQHFPGJ45]], let $I = [0, 1]$, and let $f, g \colon X \to Y$ be [[continuous maps|0VGFHTM75Y]].
A *homotopy* from $f$ to $g$ is a continuous map $H \colon X \times I \to Y$, where $X \times I$ carries the [[product topology|0GBG8B49AA]], such that $H(x, 0) = f(x)$ and $H(x, 1) = g(x)$ for all $x \in X$.

## Remarks

For each $t \in I$, the map $h_t = H(\cdot, t) \colon X \to Y$ is continuous, so a homotopy is a continuous family $(h_t)_{t \in I}$ of maps with $h_0 = f$ and $h_1 = g$.
Example: for $X = Y = \R$, $f = \id$ and $g = 0$, the map $H(x, t) = (1 - t)x$ is continuous, with $H(x, 0) = x$ and $H(x, 1) = 0$, so it is a homotopy from $f$ to $g$.
Non-example: for the same $f$ and $g$, the map $K(x, t) = x$ for $t < 1$ and $K(x, 1) = 0$ has the right ends but is not continuous at $(1, 1)$, so it is not a homotopy.
If $H$ is a homotopy from $f$ to $g$, then $(x, t) \mapsto H(x, 1 - t)$ is a homotopy from $g$ to $f$.