Which of the following statements about cardinal exponential are provable in $ZF$ The title is self explanatory, I want to see which of the below three statements are provable in $ZF$. As much as possible I will give my thoughts and reasoning.
1.)$\kappa\leq\lambda \implies \kappa^{\mu}\leq \lambda^{\mu}$
2.)$\kappa\leq\lambda \implies \mu^{\kappa}\leq \mu^{\lambda}$
3.)$\kappa < \lambda \implies \kappa^{\mu} < \lambda^{\mu}$
Firstly the 3rd statement is, in my opinion, too strong hence I crossed that one out. 
I choose 1. to be the answer as if I read it like this "there is an injection from $\kappa$ into $\lambda$, then there is an injection from $Hom(\mu,\kappa) \rightarrow Hom(\mu, \lambda)$" and I believe this is true as we can define such a function from composition.
As for 2.), I tried doing it similarly as I did for 1. but realize I can't really do it without some form of well-ordering or choice. But this isnt a good reason why 2. is false, hence if possible I would wish for a formal reason.
Thanks and cheers
 A: 2 is false when $\kappa=\mu=\varnothing$ and $\lambda\neq \varnothing$. However, if either $\kappa=\lambda=\varnothing$ and $\mu$ is arbitrary, or $\kappa=\varnothing$ and $\mu\neq\varnothing$, you can verify the implication holds. 
Assume $\kappa\neq\varnothing$.
To prove it in general, let $f\colon\kappa\to\lambda$ be an embedding. You can pick it because the set of embeddings is nonempty and you are only making one choice. Pick $x_0\in\kappa$; again, this is just one choice, so you can make it. Define $g\colon\lambda\to\kappa$ by
$$g(x) = \left\{\begin{array}{ll}
a &\text{if }x=f(a);\\
x_0 &\text{if }x\notin \mathrm{Im}(f).
\end{array}\right.$$
Note that $g$ is surjective, hence right-cancellable (this does not require choice: if $hg = kg$, then for all $a\in\kappa$ let $u\in\lambda$ with $g(u)=a$; then $h(a) = hg(u) = kg(u) = k(a)$, hence $k=h$). 
Now define a map from $\mathbf{Set}(\kappa,\mu)\to\mathbf{Set}(\lambda,\mu)$ by pre-composing with $g$: $h\longmapsto hg$. Since $g$ is right-cancellable, this map is one-to-one, hence $\mu^{\kappa}\leq \mu^{\lambda}$.
3 is false, even if you require $\mu\neq\varnothing$ (of course, if $\mu=\varnothing$ then it is false as well): $\kappa=\mu=\aleph_0$, $\lambda=2^{\aleph_0}$ give $\kappa\lt\lambda$, but $\kappa^{\mu}=2^{\aleph_0}=\lambda^{\mu}$. I believe these equalities hold in ZF since we are working specifically with alephs (in their guise as ordinals), but they certainly hold in ZFC and since this implication is provably false in ZFC, assuming the consistency of ZF you cannot prove it in ZF. 
