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If $X$ and $Y$ are topological spaces, then $X^Y$ is the set of all continuous functions from $Y$ into $X$.

There is some text in my textbook that leads me to believe this is different from $Hom(Y,X)$.

But in the category of topological spaces, isn't $Hom(Y,X)$ just the set of all continuous functions $Y \to X$? So what is the difference?

Here is a 'snippet' of the text that has lead me to the conclusion - (I can write out more if it helps)

it follows that each $F_z$ is continuous and that the target of $F^\#$ is indeed $X^Y$ (not merely $Hom(Y,X))$

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    It probably refers to the fact that $X^Y$ is endowed with a topology whereas $\operatorname{Hom}{(Y,X)}$ is only regarded as a set (or, more subtly, that two different topologies are considered). It would be helpful if you told us what textbook you're referring to.2011-04-15
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    @Theo: An Introduction to Algebraic Topology by Rotman. See my comment to Miha below - I think you are correct, that $X^Y$ is endowed with the compact-open topology2011-04-15
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    Ok, very good, then. I made the remark on the textbook only to point out that often enough the same notation is used to denote completely different things in the literature, so knowing that *some* book uses *some* notation isn't very helpful in answering a specific question.2011-04-15

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$X^Y$ usually denotes an object (called an exponential object) of the category in question. In a suitable category of topological spaces, it is the space of continuous functions $Y\to X$, equipped with the compact-open topology.

The other possibility is that $Hom(Y,X)$ is being iterpreted as the set of all functions $Y\to X$ and $X^Y$ as the set of all continuous functions.

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    I think you have it (@Theo) as well - $X^Y$ is assumed throughout to be equipped with the compact open topology.2011-04-15