Let $\kappa , \lambda$ be cardinals with $\omega \leq \lambda \leq \kappa.$
Prove $\kappa^{\lambda} = |\{X: X \subseteq \kappa, |X|=\lambda\}|$.
i.e could anyone advise me on how to construct a bijection between $\{f \mid \text{$f$ is a function},~{\rm dom}(f) = \lambda,~ {\rm ran}(f) \subseteq \kappa\}$ and $\{X: X \subseteq \kappa, |X|=\lambda\}$? Thank you.
Given a subset $X$ of $\kappa$ of size $\lambda$, fix a bijection $f_X:\lambda\to X$. Note $f_X$ is a function from $\lambda$ to $\kappa$ and the assignment $X\mapsto f_X$ is injective.
Conversely, recalling that a function is the same as its graph, any function $f:\lambda\to\kappa$ is a subset of $\lambda\times\kappa$ of size $\lambda$. But $\lambda\times\kappa$ is in bijection with $\kappa$, so we can identify $f$ with a subset $X_f$ of $\kappa$ of size $\lambda$. This assignment is also injective.
We are now done by the Bernstein-Schroeder theorem.