Let $Y^X$ denote the set of all functions $f: X \to Y$. If $X$ and $Y$ are topological spaces, let $M(X,Y)$ denote the set of all continuous maps $f: X \to Y$, endowed with the compact-open topology. Consider the functions $\varphi: Z^{X \times Y} \to (Z^Y)^X$ and $\psi: (Z^Y)^X \to Z^{X \times Y}$ defined by $\varphi(f)(x)(y) = f(x,y)$ and $\psi(g)(x,y) = g(x)(y)$, for all $f \in Z^{X \times Y}$, $g \in (Z^Y)^X$, $x \in X$ and $y \in Y$. Is easy to verify that $\varphi$ and $\psi$ are mutual inverse functions.

In Aguilar's Algebraic Topology from a Homotopical Viewpoint, we find the following results:

1.3.1. Theorem. If $X$, $Y$ and $Z$ are topological spaces with $Y$ Hausdorff and locally compact, then the image of $M(X \times Y, Z)$ under $\varphi$ is $M(X, M(Y,Z))$, whereas the image of $M(X, M(Y,Z))$ under $\psi$ is $M(X \times Y, Z)$. Therefore, the restrictions $\varphi \restriction M(X \times Y, Z)$ and $\psi \restriction M(X, M(Y,Z))$ are inverse bijections between $M(X \times Y, Z)$ and $M(X, M(Y,Z))$.

1.3.2. Theorem. If $X$, $Y$ and $Z$ are topological spaces such that $X$ and $Y$ are Hausdorff and $Y$ is locally compact, then $\varphi \restriction M(X \times Y, Z)$ is a homeomorphism onto $M(X, M(Y,Z))$.

My doubt is about the last theorem above. I cannot see in its proof where the condition of $X$ being Hausdorff come into play. Is this condition really important? Here is a picture showing this proof and its context in the book:

Aguilar's book

To confuse me further, Dugundji's Topology presents the following:

5.3 Theorem (chapter XII). If $X$, $Y$ and $Z$ are topological spaces with $Y$ Hausdorff locally compact and $X$, $Z$ arbitrary, then $\varphi \restriction M(X \times Y, Z)$ is a homeomorphism onto $M(X, M(Y,Z))$.

I have adapted the terminology and notation in this statement, but the word "arbitrary" really appears there in the book. So, is the condition of $X$ being Hausdorff dispensable? Why did Aguilar et al. include this condition in the statement of Theorem 1.3.2?

  • 1
    $\begingroup$ So topology on any $M(A, B)$ is compact-open? $\endgroup$
    – user87690
    Aug 12, 2013 at 11:06
  • $\begingroup$ Yes! $M(A,B)$ is endowed with the compact-open topology. $\endgroup$ Aug 13, 2013 at 18:49

2 Answers 2


First, there are different choices how to define local compactness outside of Hausdorff spaces. Following implications hold. If a point has neighborhood basis made of compact neighborhoods then it has some compact neighborhood. If all points have compact neighborhoods and the space is Hausdorff then it's regular. If all points have compact neighborhoods and the space is regular (not necessarily Hausdorff) then all points have neighborhood bases made of compact neighborhoods. I'll call a space locally compact iff all points have neighborhood bases made of compact neighborhoods.

For the theorem 1.3.1. $φ$ is injective function $M(X × Y, Z) \to M(X, M(Y, Z))$ for any $X, Y, Z$. If we want it to be onto, we need local compactness of $Y$ (Hausdorff is not necessary). Also note that having this bijection means that $– × Y$ is left adjoint to $M(Y, –)$.

Now for the topology on $M(A, B)$. It is compact-open. But what compact subsets do we consider? All? Hausdorff only? This matters. Dugundji considers only Hausdorff compacts and that's why he doesn't need $X$ to be Hausdorff.

For $φ$ to be continuous you don't need $Y$ to be locally compact. You need that $U^{K × L}$ is open in $M(X × Y, Z)$ and that sets $(U^L)^K$ form subbasis of $M(X, M(Y, Z))$ for $U$ open in $Y$ and $K$, $L$ considered compacts in $X$, $Y$ respectively. For the first we need $K × L$ to be considered compact. For the latter we need $K$ to be normal. (I'm not sure if a product of non-regular normal compacts is normal.) So for $φ$ to be continuous we can either consider only Hausdorff (or even regular) compacts or consider any compacts and assume that $X$ is Haussdorf (or regular).

For $φ$ to be homeomorphism we need the conditions above and $U^{K × L}$ to form a subbasis of $M(X, M(Y, Z))$. For this we need $K$, $L$ to be regular (or maybe K suffice since $Y$ is locally compact). To sum up, if we consider only Hausdorff compacts (or even only regular compacts) then the theorem holds even without Hausdorff assumption (for $X$ and for $Y$). On the other hand if we assume $X, Y$ to be Hausdorff (maybe only $X$ would suffice) then all compacts are regular so the theorem also holds.

Update: Here is a more polished statement. Let's consider all compacts in the definition of compact-open topology. Then we need $Y$ to be locally compact and $X$ to have only regular compacts. This second condition holds for both $X$ Hausdorff or $X$ regular. There is a notion of preregular space which means that any two topologically distinguishable points are separated by disjoint neighborhoods. So Hausdorff is exactly preregular + $T_0$ and regular implies preregular. Moreover, compacts in preregular space are regular. So we have the following statement: If $X$ is preregular and $Y$ is locally compact (not necessarily Hausdorff), then $φ: M(X × Y, Z) \to M(X, M(Y, Z))$ is a homeomorphism.


The assumption that $X$ is Hausdorff is already used in the first step of the displayed proof—in the claim that the sets of the form “$(U^L)^K$” form a subbasis. Said claim follows from Lemma~XII.5.1 (a) of Dugundji’s Topology. Alternatively, see Tammo tom Dieck's Topologie, Satz 11.1.

Note that, in Topology, Dugundji implicitly assumes all spaces to be Hausdorff! This is mentioned in the introduction. There is also a reminder at the very beginning of section XII.6.

That said, I do not know whether the assumption on $X$ is essential or whether it just makes for a more convenient proof.


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