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In Switzer's book, it is defined (in page 3):

The exponential law: if $X, Z$ are Hausdorff spaces and $Z$ is locally compact, then the natural function $$E:Y^{Z\times X}\to (Y^Z)^X$$ defined by $(Ef(x))(z)=f(z,x)$ is a homeomorphism.

I am a bit puzzled by "$(Ef(x))(z)=f(z,x)$", as it does not seem to make sense with $E:Y^{Z\times X}\to (Y^Z)^X$.

Since $Y^{Z\times X}$ is the set of maps $f: Z\times X\to Y$, shouldn't $E$ take in a function $f:Z\times X\to Y$, and produce a function $g: X\to Y^Z$?

Secondly, in the definition, firstly $f$ is a single variable function of $x$ and then later $f$ is a 2-variable function of $(z,x)$ which confuses me.

Thanks for any help!

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    The book exactly states that. $g=Ef$ is a function from $X$ to $Y^Z$. When you apply $g$ on some $x$, you get function $g(x)=(Ef)(x)$ from $Z$ to $Y$. – 2017-01-13
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    Yes indeed! I see it now. Thanks. – 2017-01-13
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    @user251257 You should make that into an answer... – 2017-01-13

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You want to know the value of $E$ on $f: Z \times X \rightarrow Y$. This $E(f)$ must be a function from $X$ to $Y^Z$, so must specify what the value $E(f)(x)$ is for some arbitary $x \in X$; namely some function from $Z$ to $Y$, and so we must also specify what $E(f)(x)(z)$ for any $z \in Z$ is. The answer forces itself, really: we have $f$ and some points $x$, $z$, so we take the obvious $f(z,x)$ which is indeed a point of $Y$, as needed. So this is well-defined, and in some way obvious (it's often called a canonical map).