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First let $\Lambda$ be the bijective mapping between $Y^{Z \times X}$ and $(Y^X)^Z$ defined as follows: every mapping $f: Z \times X \to Y$ defines a set of mappings from $X$ to $Y$: for each $z \in Z$ is $f_z:X \to Y$ defined as $f_z(x) = f(z,x)$. The mapping $z \mapsto f_z$ of $Z$ to $Y^X$ obtained this way we denote as $\Lambda(f)$.

Engelking calls this the exponential mapping.

Then he goes on to define a topology on $C(X, Y)$ called the pointwise topology as the restriction of the product topology on $Y^X$ restricted to $C(X,Y)$. We can also see that this is equal to the topology generated by the subbasis $$\{M(x, U) : x \in X \text{ and $U$ open in $Y$}\},$$ where $M(x,U) := \{f \in C(X,Y) : f(x) \in U\}$.

So now I can finally ask the question I want to ask...

Give $C(X,Y)$, $C(Z \times X, Y)$ and $C(Z, C(X, Y))$ the pointwise topology.

How do I now show that $\Lambda:C(Z \times X, Y) \to C(Z, C(X,Y))$ is an embedding? I'm drowning in a syntax mess. I don't need a full solution a road map is fine.

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up vote 5 down vote accepted

I'll show continuity. The pointwise topology on $C(X,Y)$ is precisely what it says: a net $g_i$ converges in it, $g_i\to g$, if and only if for all $x\in X$ we have $g_i(x)\to g(x)$ in $Y$. (Indeed, this is because a net in a product converges iff all projections converge.)

So suppose $f_i\to f$ in $C(Z\times X,Y)$. We need to show $\Lambda (f_i)\to \Lambda (f)$. The latter means that, for all $z\in Z$, we have $(f_i)_z\to f_z$ in $C(X,Y)$ . But this means that for all $z\in Z$, we have that for all $x\in X$, $(f_i)_z(x)\to f_z(x)$ in $Y$. In other words, it means that for all $(z,x)\in Z\times X$, we have $f_i(z,x)\to f(z,x)$ in $Y$. But this is precisely our assumption $f_i\to f$ in $C(Z\times X,Y)$.

So it's basically a tautology: the maps are defined by 'currying' (holding the function pointwise fixed), and we are considering the pointwise topology.

To show that in fact $\Lambda(f)$ is continuous for continuous $f$, a similar argument can be used (namely $\Lambda(f)(z_i)\to \Lambda(f)(z)$ for nets $z_i\to z$), and the same for the well-definedness and continuity of the inverse.

(I think this is one of the places where using nets is superior to working directly with open set / subbases of the topology.)

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I think that in this way we can show that $Y^{Z\times X}$ and $(Y^X)^Z$ are homeomorphic. (Continuouiy of both $\Lambda$ and $\Lambda^{-1}$ can be shown via nets.) If we work with the restriction of these topologies to spaces of continuous map, it only remains to show that every element of $C(Z\times X,Y)$ is mapped to an element of $C(Z,C(X,Y))$. (Again, nets should be useful for this.) – Martin Sleziak Jul 2 '11 at 19:24
Great. Since I'm doing topology I'll take filters instead of nets then at least there is something left to think about for me ;-). – Jonas Teuwen Jul 2 '11 at 19:28
For the record: Doesn't imbedding in topology just mean that the mapping in question is a homeomorphism onto its range? In this case I already know what $\Lambda$ and its inverse are. – Jonas Teuwen Jul 2 '11 at 19:38
@Jonas, yes that is correct. – wildildildlife Jul 2 '11 at 19:46

For the pointwise topology, I don't know. For the compact-open topology, with hypotheses on the spaces, the map you are talking about can be a homeomorphism and you can find it in the book of Maunder. For the pointwise topology, there are results in the book of Munkres.

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I am sorry, but 'there are result in Munkres' does not seem really helpful. – wildildildlife Jul 2 '11 at 19:23

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