# What is the topology on $[0,1]^2$ in the defintion of a homotopy?

I have a question about the definition of a homotopy between loops:

Let $\alpha$ and $\beta$ be loops with base point $x$ in a topological space $X$. A homotopy from $\alpha$ to $\beta$ is a continuous function $H$ from $[0,1]^2$ to $X$ such that $H(s,t)=f_s(t)$ where $f_s$ is a loop with base point $x$ and such that $f_0=\alpha$ and $f_1=\beta$.

What are the opens relative to $[0,1]^2$? If we're calling $H$ continuous, don't we need to know what topology on $[0,1]^2$ we're talking about? Is a certain topology implied?

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Given any two spaces, $X$ and $Y$, there is a unique way to define a topology on $X \times Y$ so that continuous maps from a space to $X \times Y$ are all given by pairs $(f,g)$ where $f$ is a continuous map to $X$ and $g$ is a continuous map to $Y$. One way to define it is as the topology with the fewest open sets so that the projection maps onto the factors are continuous. In particular, this puts a topology on $[0,1]^2$. :) – Dylan Wilson Dec 12 '12 at 20:51
The good ol' topology that is induced on $[0,1]^2$ from $\mathbb{R}^2$. – levap Dec 12 '12 at 20:51
(also levap's comment is way more useful in this particular space, but it's good to know the other thing if you plan on doing anything at all with homotopy!) – Dylan Wilson Dec 12 '12 at 20:52
@DylanWilson Actually, not only that all functions to $X\times Y$ are of this form, but that any pairs of continous functions $f:Z\to X,g:Z\to Y$ yields such a continuous function. – Thomas Andrews Dec 12 '12 at 21:07

It's just the usual topology, i.e. the product topology on $[0,1] \times [0,1]$.