Exercise 9.1.10

Proof. First, we clearly have that

So let $A$ and $B$ be sets such that $\left|A\right|=\kappa$ and $\left|B\right|=\lambda$. Then by definition we have $\kappa =\left|A\right|$ and ${\kappa }^{\lambda }=\left|A^B\right|$. We also have that there is a bijective $g:\lambda \to B$ since $\left|\lambda \right|=\lambda =\left|B\right|$. We construct a bijection $f$ from $A^B$ to $A$. So, for any $h\in A^B$ let $f(h)=h\circ g$. Clearly, since $g:\lambda \to B$ and $h:B\to A$, it follows that $f(h):\lambda \to A$ and hence clearly $f(h)\in A$ (since $f(h)(\alpha )\in A$ for each $\alpha <\lambda$) so that $f$ is a function from $A^B$ to $A$.

We also claim that $f$ is injective. So consider any $h_1$ and $h_2$ in $A^B$ where $h_1\ne h_2$. It then follows that there must be a $b\in B$ where $h_1(b)\ne h_2(b)$. Then let $\alpha =g^{-1}(b)$, noting that $g^{-1}$ is a bijective function from $B$ to $\lambda$ since $g$ is bijective. We then have that

so that $f(h_1)\ne f(h_2)$. Since $h_1$ and $h_2$ were arbitrary this shows that $f$ is indeed injective.

We also claim that $f$ is onto. So consider any $h^{\prime }\in A$ so that $h^{\prime }$ is a function from $\lambda$ to $A$ (since $h^{\prime }(\alpha )\in A$ for each $\alpha <\lambda$). Then let $h=h^{\prime }\circ g^{-1}$ so that clearly $h$ is a function from $B$ to $A$ since $g^{-1}:B\to \lambda$ and $h^{\prime }:\lambda \to A$. We then have that

where $i_{\lambda }$ is the identity function from $\lambda$ to $\lambda$. Since $h^{\prime }$ was arbitrary, this shows that $f$ is indeed onto.

Thus, since we have shown that $f$ is a bijection, it follows that

as desired. □