# Does the bijection of function sets imply bijection of sets?

Let $X,Y$ and $Z$ be sets. Is it true that if the set of functions from $Z$ to $X$ is in bijection to the set of functions from $Z$ to $Y$, then $X$ is in bijection to $Y$? Or are there any subtleties with axioms of set theory? If not, how do you prove it?

-

There is a bijection between functions from $\mathbb N$ to $\{1,2\}$ and functions from $\mathbb N$ to $\{1,2,3,4\}$.

-
Hmm. But what about the consequence of the Yoneda Lemma: $\text{Hom}_{\mathcal{C}}(A,B)\cong \text{Hom}_{\mathcal{C}}(A,C)\Rightarrow B\cong C$? What do I not get here?...Why does it not hold in the category of sets? –  PatrickMcGill Mar 13 '14 at 7:14
You need a natural bijection. –  Zhen Lin Mar 13 '14 at 8:33
Ah..natural in all variables? And then, how is the result proved for sets? –  PatrickMcGill Mar 13 '14 at 10:57
It's proved with a bijection between $\mathbb{N}$ and $\mathbb{N}^2$. $\;$ –  Ricky Demer Mar 13 '14 at 11:58
What?! I mean the result in my comment... –  PatrickMcGill Mar 13 '14 at 12:05

Your question is not soft, even if you tagged it that way. It is on the other end slightly ambiguously stated. So let me restate it precisely: If, for every set $Z$, the sets $X^Z \cong Y^Z$ , are the sets $X$ and $Y$ isomorphic?

The answer is yes. Just take as $Z$ the singleton set $*$. Then $$Y \cong Y^* \cong X^* \cong X$$

-
Thanks! Is this the proof of the fact that the set $Y^X$ is the unique exponential for the sets $X$ and $Y$? And more importantly, is this the closest to uniqueness that the function set $Y^X$ gets in this context? I mean, in the statement "$Y^X$ is the exponential of $X$ and $Y$", what does "ïs" mean? Thanks again! –  PatrickMcGill Mar 13 '14 at 14:49