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For context: an exponential of objects $B$ and $A$ in a category $\mathcal{C}$ is defined as an object $B^A$ and a morphism $\epsilon: B^A \times A \rightarrow B$ such that for every object $C$ in $\mathcal{C}$ and every morphism $f: C \times A \rightarrow B$ there is a unique morphism $\lambda f: C \rightarrow B^A$ such that $\epsilon \circ (\lambda f \times \mathrm{id}_A) = f$. The $\lambda$ operation in this context is often referred to as currying, or partial application.

If $C=1$ (the terminal object in $C$), we get a bijection between morphisms $A \rightarrow B$ and morphisms $1 \rightarrow B^A$, which allows us to interpret the exponential object as the "space of morphisms" from $A$ to $B$ (aka "internal hom").

The question is, why is it crucial to require this bijection for every object $C$ instead of only for the terminal object? I suppose the definitions are not equivalent (admit I didn't check, though); the question is more in the lines of "why is this definition more interesting/important/natural/obvious/etc.?".

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If you want to interpret $B^A$ as a "space of morphisms" you need to know how to map into or out of this space. The definition of exponential object tells you how to map into this space; that is, it tells you what a $C$-parameterized family of maps $A \to B$ is, namely it's a map $C \times A \to B$. This uniquely determines $B^A$, if it exists, by the Yoneda lemma, because it gives a universal property that $B^A$ satisfies. If you don't require a condition for every $C$ then you don't have a universal property and so in particular no guarantee of uniqueness.

Already the definition of $A \times B$ does not refer just to points $1 \to A \times B$ but to arbitrary maps $C \to A \times B$, the point being that the universal property of the product tells you not only what a point in $A \times B$ is but what a $C$-parameterized family of points is (roughly speaking).

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  • $\begingroup$ So if I wanted to illustrate the point, I would look for two non-isomorphic (or not uniquely isomorphic) objects that fulfill the weaker property, and thus show that I failed to capture the notion of "the" morphism space? $\endgroup$ – askyle Apr 26 '14 at 15:44
  • $\begingroup$ OK, I played with it a bit and I see the point about universality now. Didn't have to use Yoneda to show universality, though... $\endgroup$ – askyle Apr 27 '14 at 11:14
  • $\begingroup$ @askyle: yes. It's easy to construct a counterexample in the category of $G$-sets for $G$ a nontrivial group. Yoneda is used to show that universal objects are unique up to unique isomorphism (by which I mean that representing objects of functors are unique up to unique isomorphism). $\endgroup$ – Qiaochu Yuan Apr 27 '14 at 17:37
  • $\begingroup$ Can I use Yoneda here even if $\mathcal{C}$ is not locally small? $\endgroup$ – askyle Apr 28 '14 at 11:46
  • $\begingroup$ @askyle: not without more work (e.g. maybe one can pick a universe $U$ such that $C$ is locally $U$-small or something...). I usually try to avoid categories that aren't locally small. $\endgroup$ – Qiaochu Yuan Apr 28 '14 at 13:09
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Currying has bijection: $\mathsf{Hom}(A\times B,C)\cong \mathsf{Hom}(A,C^B)$

This bijection comes from product-hom adjunction.

You get evaluation $\epsilon$ by $A=C^B$. Evaluation wouldnt work if you do $A=1$. Both need to work. Note that to get evaluation working, you need identity function $\mathrm{id}_{C^B}$, this is needed in $\mathsf{Hom}(C^B \times B,C)\cong \mathsf{Hom}(C^B,C^B)$. Identity function $\mathrm{id}_K$ requires that $K$ is an object, so $C^B$ is an object, instead of arrow.

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