I'm trying to prove the equipotency of these two sets. I know I have to find an invertible function $f:(X^Y)^Z\rightarrow X^{Y\times Z}$ between the two sets.
Which means I have to find a function $g:X^{Y\times Z}\rightarrow (X^Y)^Z$ such that: \begin{align} f\circ g (h)&=h \quad \text{for any } h\in X^{Y\times Z}\quad \text{and}\\ g\circ f (k)&=k \quad \text{for any } k\in (X^Y)^Z \end{align} Now I can note that \begin{align} h:Y\times Z&\longrightarrow X\\ (y,z)&\longmapsto h(y,z)\\ \\ k:Z&\longrightarrow X^Y\\ z&\longmapsto k(z):Y\rightarrow X\\ &\qquad\qquad\quad\text{ } y\mapsto k(z)(y) \end{align} Does this help me in any way?
I'm clueless as to what to pay attention to in order to reach that, any help is much appreciated!
The maps $f\colon (X^Y)^Z\to X^{Y\times Z}$ and $g\colon X^{Y\times Z}\to (X^Y)^Z$ may look somewhat convoluted, but are after all natural (also in the technical sense): If $\phi\in (X^Y)^Z$, i.e., $\phi$ is a map $Z\to X^Y$, then for every $z\in Z$, $\phi(z)$ is an element of $X^Y$, i.e., a map $Y\to X$. In order to define $f(\phi)$, we need to exhibit a map $Y\times Z\to X$. The - indeed natural - candidate is given by $ \langle y,z\rangle\mapsto \phi(z)(y).$ So to summarize, $$ f\colon \phi\mapsto (\langle y,z\rangle\mapsto \phi(z)(y))$$ Similarly, given $\psi\in X^{Y\times Z}$, i.e., $\psi\colon Y\times Z\to X$, $g$ shall map this to a map from $Z$ to the set of maps from $Y$ to $X$. The -again natural - candidate is to let $g(\psi)$ map $z$ to the map $Y\to X$ given by $y\mapsto \psi(\langle y,z\rangle)$. To summrize again, $$ g\colon \psi\mapsto (z\mapsto(y\mapsto \psi(\langle y,z\rangle))).$$ As the definitions of $f,g$ were natural, we expect $f\circ g$ and $g\circ f$ to be natural as well - and what could be more natural as map from $(X^Y)^Z$ (or $X^{Y\times Z}$) to itself than the identity? Wll, we have to check nevertheless: For $\phi\in (X^Y)^Z$, $$ \begin{align}(g\circ f)(\phi)&=g(f(\phi))\\&=g(\underbrace{\langle y,z\rangle\mapsto\phi(z)(y)}_{=:\psi})\\ &=(z\mapsto (y\mapsto \psi(\langle y,z\rangle)))\\ &=(z\mapsto(y\mapsto \phi(z)(y)))\\ &=(z\mapsto \phi(z))\\&=\phi\end{align}$$ and vice versa for $\psi\in X^{Y\times Z}$, $$\begin{align}(f\circ g)(\psi) &=f(g(\psi))\\ &=f(\underbrace{z\mapsto(\underbrace{y\mapsto\psi(\langle y,z\rangle)}_{=:\chi})}_{=:\phi})\\ &=(\langle y,z\rangle\mapsto \phi(z)(y))\\ &=(\langle y,z\rangle\mapsto \chi(y)\\ &=(\langle y,z\rangle\mapsto \psi(\langle y,z\rangle))\\ &=\psi\end{align} $$