For $A$ an infinite set, and $f: A^A \to A$, does there exist $a\in A$ such that $f^{-1}(\{a\}) \cong A^A$? I've been wondering about this question, and unfortunately I do not have much experience with set theory.
The notation: $A^A := \{g:A\to A\}$.
The reason I suspect this is very weak actually, and that is that for finite sets $B$, there exists a point $b\in B$ such that the preimage $|f^{-1}(\{b\})|\geq |B|^{|B|-1}$, and if I would just stupidly substitute the formula for infinite sets, it would yield a bijection between the preimage of $b$ and $B^B$.
Thanks in advance!

Yes.
König's theorem states that the cofinality of $2^\kappa$ is strictly greater than $\kappa$. This means that if we partition $2^\kappa$ into at most $\kappa$ parts, one of them is necessarily of size $2^\kappa$. 
To see why, note that you can formalize cofinality by stating that the cofinality of an infinite cardinal $\lambda$ is the least $\mu$, such that we can partition $\lambda$ into $\mu$ parts, all of which are less than $\lambda$ in size.
Of course, the axiom of choice is needed here. Indeed without it, for example, it is consistent that $\Bbb{N^N}$ is the countable union of countable sets, and that gives us a map from $\Bbb{N^N\to N}$ where every fiber has size $\aleph_0$, but the entire set has size $2^{\aleph_0}$.
