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**MachinePL1993** I need to proof knowing that some two sets $\displaystyle A_{1}$ and $\displaystyle A_{2}$ have the same cardinality that sets $\displaystyle A_{1}^{B}$ and $\displaystyle A_{2}^{B}$, which are sets of maps from some set $\displaystyle B$ to sets $\displaystyle A_{1}$ and $\displaystyle A_{2}$ respectively, also have the same cardinality.

So here's my attempt:

So we know that those two sets are equal, so that means that there's a bijection between them, like this:

$\displaystyle f:A_{1} \rightarrow A_{2}$

I have to create a bijection between $\displaystyle A_{1}^{B}$ and $\displaystyle A_{2}^{B}$ to prove that they have the same cardinality. So let $\displaystyle F$ be this bijection.

$\displaystyle F$ is supposed to take a function that goes from $\displaystyle B$ to $\displaystyle A_{1}$ and return a function that goes from $\displaystyle B$ to $\displaystyle A_{2}$. $\displaystyle f_{1}:B \rightarrow A_{1} $ for some $\displaystyle b \in B$ equals $\displaystyle x \in A_{1}$ meaning $\displaystyle f_{1}(b)=x$.

Let us construct a function $\displaystyle f2:B \rightarrow A_{2} $ that does this $\displaystyle f_{2}(b)=f(f_{1}(b))=f(x)$. This function takes $\displaystyle b$ from $\displaystyle B $ and returns an element in $\displaystyle A_{2}$ by assigning one to the result of $\displaystyle f_{1} $ by the bijection between sets $\displaystyle A_{1}$ and $\displaystyle A_{2}$.

So in the end, a bijection between sets $\displaystyle A_{1}^{B}$ and $\displaystyle A_{2}^{B}$ is a function $\displaystyle F(f_{1})=f_{2}$.