TITLE: Why is currying part of the definition of exponential objects?
QUESTION [5 upvotes]: 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.?".

REPLY [2 votes]: 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.