Suppose $X$ and $Y$ denote objects of $\mathrm{Set}$. Is there any reason not to say: therefore, $\mathrm{Hom}(X,Y)$ is itself an object of $\mathrm{Set}$? Furthermore, is there any reason not to write $Y^X = \mathrm{Hom}(X,Y)$ and/or $Y^X \cong \mathrm{Hom}(X,Y)$?

A good answer should probably mention the following concepts.

  1. The internal/external distinction.
  2. Exponential objects; and what, if anything, is the difference between $\mathrm{Hom}(X,Y)$ and $Y^X,$ (specifically in the case where $X$ and $Y$ are objects of $\mathrm{Set}$.)
  3. The principle of equivalence.
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    $\begingroup$ I don't get the question. It is a set, hence an object of the category of sets. Is there anything to discuss here?! $\endgroup$ Jan 17 '14 at 21:43
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    $\begingroup$ @MartinBrandenburg, so you're saying that writing $Y^X = \mathrm{Hom}(X,Y)$ is acceptable, in your opinion? $\endgroup$ Jan 17 '14 at 21:45
  • $\begingroup$ I think the notation $Y^X$ or $(-)^X:\mathcal C\to \mathcal C, Y\mapsto Y^X$ is used in general if this is a functor which is right adjoint to the product functor $(-)×X: Z\mapsto Z×X$. In $\mathbf{Set}$ if $Y^X=$Hom$(X,Y)$ then this is the case. $\endgroup$ Jan 17 '14 at 21:50
  • $\begingroup$ This is the trivial case of "enriched category". In other words, a Set-enriched category is nothing but an ordinary category. $\endgroup$
    – user314
    Jan 18 '14 at 14:31

$\mathrm{Hom}(X,Y)$ is a set. Together with the evaluation map ($\mathrm{Hom}(X,Y) \times X \to Y$), it is also an exponential object, because it satisfies the definition of one. However, a consequence of the definition of exponential objects is that, if they exist, they are unique up to (unique) isomorphism (not equal, but more than just isomorphic). So we tend to write $Y^X$ to mean any (fixed) object (together with an implicitly defined evaluation map) of this form, and say words like "the exponential object", understanding implicitly that there are many, but they are all naturally isomorphic, so all have the same properties.

  • $\begingroup$ Okay. I'm satisfied. $\endgroup$ Jan 17 '14 at 23:55
  • $\begingroup$ It suddenly occurs to me that if we want $\mathrm{Hom}$ to be a functor $\mathrm{Set}^\mathrm{op} \times \mathrm{Set} \rightarrow \mathrm{Set}$, we'd better have $\mathrm{Hom}(X,Y)$ denote an object of $\mathrm{Set}$. $\endgroup$ Jan 18 '14 at 1:15

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