Homotopy

Definition
We begin by defining homotopies that take time $$1$$, but in the last subsection consider a variant notion of a homotopy that could take time $$T > 0$$.

Definition as a jointly continuous map from the product with the unit interval
Suppose $$X, Y$$ are topological spaces and $$f,g:X \to Y$$ are continuous maps from $$X$$ to $$Y$$. Let $$I$$ be the defining ingredient::closed unit interval $$[0,1]$$.

A defining ingredient::jointly continuous map $$F: X \times I \to Y$$ is termed a homotopy from $$f$$ to $$g$$ if for every $$x \in X$$, $$F(x,0) = f(x)$$ and $$F(x,1) = g(x)$$.

Note that $$F$$ has to be a continuous map from $$X \times I$$ equipped with the product topology. It is not sufficient to require that $$F$$ be a separately continuous map in each coordinate, i.e., it is not enough to insist that $$x \mapsto F(x,t)$$ is continuous for each $$t$$ and $$t \mapsto F(x,t)$$ is continuous for each $$x$$.

Definition as a path in a function space
This definition works (at least) in the case that both $$X$$ and $$Y$$ are compactly generated Hausdorff spaces (probably in more cases). Under this definition, a homotopy between two continuous maps $$f,g:X \to Y$$ is a path from the point $$f$$ to the point $$g$$ in the topological space $$C(X,Y)$$ defined as the set of continuous maps from $$X$$ to $$Y$$ equipped with the compact-open topology.

Equivalence of definitions
A map $$F: X \times [0,1] \to Y$$ is equivalent to a map $$\gamma_F$$ from $$[0,1]$$ to the space $$Y^X$$ of functions from $$X$$ to $$Y$$, via the following rule:

$$\! \gamma_F(t) := x \mapsto F(x,t)$$

and in reverse:

$$\! F_\gamma(x,t) := (\gamma(t))(x)$$

It is further true that if $$F$$ is jointly continuous, then for each $$t \in [0,1]$$, $$\gamma_F(t)$$ is a continuous map. Thus, the homotopy from $$f$$ to $$g$$ is a map $$\gamma$$ from $$[0,1]$$ to the set $$C(X,Y)$$ of all continuous maps from $$X$$ to $$Y$$, where $$\gamma_R(0) = f$$ and $$\gamma_F(1) = g$$.

However, any set map from $$[0,1]$$ to $$C(X,Y)$$ need not be a homotopy, because the corresponding map $$F$$ need not be jointly continuous. It turns out that when both $$X$$ and $$Y$$ are compactly generated Hausdorff spaces, then a map $$\gamma:[0,1] \to C(X,Y)$$ is a continuous map to $$C(X,Y)$$ equipped with the compact-open topology iff the corresponding $$F_\gamma$$ is a jointly continuous map.

Variant: a homotopy that takes time $$T > 0$$
Suppose $$X, Y$$ are topological spaces and $$f,g:X \to Y$$ are continuous maps from $$X$$ to $$Y$$. A homotopy from $$f$$ to $$g$$ that takes time $$T$$ is a continuous map $$F: X \times [0,T] \to Y$$ such that $$F(x,0) = f(x)$$ and $$F(x,T) = g(x)$$ for all $$x$$ in $$X$$.

Given any homotopy that takes time $$T$$, there is a linear scaling of the homotopy to a homotopy that takes time $$1$$, which would make it a homotopy in the first sense. The main advantage of considering homotopies that take time $$T$$ is that these have an associative multiplication.

Related notions

 * Self-homotopy. Also check out Category:Properties of self-homotopies
 * Smooth homotopy and piecewise smooth homotopy
 * Linear homotopy and piecewise linear homotopy