Let $X,Y,Z$ be nonempty sets. Then $\left |\left(X^Y\right)^Z\right|=\left|X^{Y\times Z}\right|$.
Please help me verify this proof! Thank you so much!
My attempt:
We define a mapping $F$ that assigns each $f\in \left(X^Y\right)^Z$ to each $g\in X^{Y\times Z}$ by $$\forall (y,z)\in Y\times Z:g(y,z):=f(z)(y)$$
- $F$ is surjective
For $g\in X^{Y\times Z}$, we define $f$ by $f(z)(y):=g(y,z)$ for all $z\in Z$ and $y\in Y$.
By definition of $F$, $F(f)=g$.
- $F$ is injective
Assume $f_1,f_2\in \left(X^Y\right)^Z$ and $F(f_1)=g_1=g_2=F(f_2)$.
$g_1=g_2\implies [\forall (y,z)\in Y\times Z][g_1(y,z)=g_2(y,z)] \implies [\forall (y,z)\in Y\times Z][f_1(z)(y)=f_2(z)(y)] \implies (\forall z\in Z)[(\forall y\in Y)(f_1(z)(y)=f_2(z)(y))] \implies (\forall z\in Z)[f_1(z)=f_2(z)] \implies f_1=f_2$.
To sum up: $F$ is bijective and thus $\left |\left(X^Y\right)^Z\right|=\left|X^{Y\times Z}\right|$.
It looks okay, but the definition of your "function" is a bit weird, correct me I am wrong but you are trying to say:
$$F:X^{Y\times Z}\to\left(X^Y\right)^Z\\((F(g))(x))(y)=g(y,x)$$ In other words, $F$ is a function who gets a function $g:Y\times Z\to X$ and return a function $F(g):Z\to X^Y$, this function gets an element from $x\in Z$ and return a function from $F(g)(x):Y\to X$ and this function gets an element from $y\in Y$ and return $((F(g))(x))(y)\in X$.
If this is what you mean then yes, the proof is okay but note that we can define inverse function to prove this easily:$$G:\left(X^Y\right)^Z\to X^{Y\times Z}\\(G(f))(x,y)=(f(y))(x)$$