Exercise 9.1.11

Proof. First suppose that $⟨A_i\mid i\in I⟩$ are sets where $\left|A_i\right|={\kappa }_i$ for all $i\in I$. Also suppose that $B$ is a set such that $\left|B\right|=\lambda$. It then follows that

and

We then aim to construct a bijection $F$ from ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i^B$ to ${\left({\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i\right)}^B$, which clearly would show the result since we would have by the above that

So suppose that $f\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i^B$ so that $f_i=f(i)\in A_i^B$ for every $i\in I$. Then define a function $g$ such that, for any $b\in B$, $g(b)=⟨f_i(b)\mid i\in I⟩$. We then have that, for any $i\in I$, $f_i(b)\in A_i$ since $f_i$ is a function from $B$ to $A_i$. Therefore $g(b)\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$, and hence $g$ is a function from $B$ to ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ so that $g\in {\left({\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i\right)}^B$. Naturally then we set $F(f)=g$ so that $F$ is indeed a function from ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i^B$ to ${\left({\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i\right)}^B$.

To show that $F$ is injective consider any $f$ and $f^{\prime }$ in ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i^B$ where $f\ne f^{\prime }$. It then follows that there is an $i_0\in I$ such that $f_{i_0}=f(i_0)\ne f^{\prime }(i_0)=f_{i_0}^{\prime }$. Then, since $f_{i_0}$ and $f_{i_0}^{\prime }$ are both in $A_{i_0}^B$, it follows that they are both functions from $B$ to $A_{i_0}$. Since $f_{i_0}\ne f_{i_0}^{\prime }$ we have that there is a $b\in B$ such that $f_{i_0}(b)\ne f_{i_0}^{\prime }(b)$. We then have that $g(b)=⟨f_i(b)\mid i\in I⟩\ne ⟨f_i^{\prime }(b)\mid i\in I⟩=g^{\prime }(b)$ since $f_{i_0}(b)\ne f_{i_0}^{\prime }(b)$ and $i_0\in I$. Therefore $F(f)=g\ne g^{\prime }=F(f^{\prime })$, which shows that $F$ is injective since $f$ and $f^{\prime }$ were arbitrary.

To show that $F$ is also onto, consider any $g^{\prime }\in {\left({\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i\right)}^B$ so that $g^{\prime }$ is a function from $B$ to ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$. Therefore, for any $b\in B$, $g^{\prime }(b)$ is a function on $I$ such that $g^{\prime }(b)(i)\in A_i$ for each $i\in I$. Clearly we have $g^{\prime }(b)=⟨g^{\prime }(b)(i)\mid i\in I⟩$. Then, for each $i\in I$, we define a function $f_i$ on $B$ such that $f_i(b)=g^{\prime }(b)(i)$. Then, since $f_i(b)=g^{\prime }(b)(i)\in A_i$, clearly each $f_i$ is a function from $B$ to $A_i$. Therefore $f_i\in A_i^B$. We then set $f=⟨f_i\mid i\in I⟩$ so that clearly $f\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i^B$. Now, we then set $g=F(f)$ so that, by the definition of $F$, $g$ is a function on $B$ such $g(b)=⟨f_i(b)\mid i\in I⟩$ for any $b\in B$. We then have that $g(b)=⟨f_i(b)\mid i\in I⟩=⟨g^{\prime }(b)(i)\mid i\in I⟩=g^{\prime }(b)$ by the definition of each $f_i$. Since $b\in B$ was arbitrary, this shows that $F(f)=g=g^{\prime }$. Thus $F$ is onto since $g^{\prime }$ was arbitrary.

We have shown that $F$ is a bijection so that the result follows as described above. □