$\mathrm{DISCLAIMER~:~}$I am not interested in working with compactly generated spaces.

This post is related to this one : Exponential Law for based spaces. I learned about the exponential law for topological spaces quite some time ago, and I thought I understood it well until I decided to reprove it today as I have been using it lately.

What confuses me is that I 'seem' to have proven it with weaker conditions than those stated in the textbooks. To be precise, I think I have shown that there is a natural homeomorphism $$\mathrm{Map}(X\times Y,Z)\simeq\mathrm{Map}(X,\mathrm{Map}(Y,Z))$$ where $Z$ is any topological space, $X$ is Hausdorff and $Y$ locally compact $no$ $Hausdorff$ $condition$ $required\dots$ In all textbooks I 'm familiar with, none of which feature a proof of the above fact, the extra assumption is made that $Y$ be Hausdorff. The proof I gave is, I think, the one I learnt in Switzer's book (if I remember right) yet I see no need for Hausdorffness in $Y$.

$\mathrm{QUESTION~1:~}$Is $Y$ Hausdorff really necessary?

Also, the reference I am currently using, Algebraic Topology from the Homotopical Viewpoint [Aguilar, Gitler, Prieto, Springer Universitext], exercice $1.3.4$ asks to show that for $X,Y,Z$ topological spaces with $X$ and $Y$ locally compact Hausdorff spaces, composition $$\mathrm{Map}(X,Y)\times\mathrm{Map}(Y,Z)\rightarrow\mathrm{Map}(X,Z),(f,g)\mapsto g\circ f$$ is continuous. Yet I'm pretty sure all you need is for $Y$ to be locally compact$\dots$

$\mathrm{QUESTION~2:~}$ Are all these extra conditions necessary?

  • $\begingroup$ Without Hausdorff, please specify what you mean by locally compact: every point has a base of open sets with compact closure, every point has a base of compact (not open) neighbourhoods, or instead of base you want just one open set with compact closure, or one compact neighbourhood? These are only equivalent under Hausdorff-like conditions... $\endgroup$ Jul 4, 2011 at 18:17
  • $\begingroup$ I used following definition : every point has a base of compact neighborhoods. $\endgroup$ Jul 4, 2011 at 18:26
  • $\begingroup$ Ok, so not necessarily compact --> locally compact. $\endgroup$ Jul 4, 2011 at 18:31
  • $\begingroup$ Yes, this is no longer true (a priori). $\endgroup$ Jul 4, 2011 at 18:36
  • $\begingroup$ Engelking states this homeomorphism for: any space Y, Z Hausdorff, X locally compact (which includes Hausdorff with him). And for any X,Z Hausdorff, any Y we have an embedding (which need not be onto). Also, the composition map is continuous in the compact open topologies for any X,Z, locally compact Y. $\endgroup$ Jul 4, 2011 at 18:45

2 Answers 2


As to Question 2, let $\Sigma$ be the composition map $\mathrm{Map}(X,Y)\times\mathrm{Map}(Y,Z)\rightarrow\mathrm{Map}(X,Z)$.

All functions spaces have the compact-open topology, with as a subbase all sets of the form $\mathrm{M}(C,U) = \{ f: f[C] \subset U \}$, where $C$ is a compact subset of the domain, and $U$ an open subset of the co-domain.

Let $(f,g)$ be a point in $\Sigma^{-1}[\mathrm{M}(C,U)]$, with $C \subset X$ compact and $U \subset Z $ open, and we want to show it's an interior point. We have by definition $(g \circ f)[C] \subset U$, or $f[C] \subset g^{-1}[U]$. As $g^{-1}[U]$ is open, and $f[C]$ is compact, both in $Y$, and if we assume $Y$ is locally compact (in the sense from the comments), we can find for each $y \in f[C]$ a compact neighbourhood $K_y$ that sits inside $g^{-1}[U]$, and so $f[C]$ is covered by finitely many sets of the form $\mathrm{Int}(K_y)$, say $\mathrm{Int}(K_{y_i})$ for $i=1 \ldots n$ and set $W$ to be the finite union of these. Then $W \subset \mathrm{Int}(\cup_{i=1}^n K_{y_i}) \subset K:=\cup_{i=1}^n K_{y_i} \subset g^{-1}[U]$ and so $\Sigma[\mathrm{M}(C,W) \times \mathrm{M}(K,U) ] \subset \mathrm{M}(C,U)$ and so we are done, as $(f,g)$ is in $\mathrm{M}(C,W) \times \mathrm{M}(K,U)$.

This convinces me that indeed we only need local compactness (in a rather strong sense) on $Y$ to show continuity of $\Sigma$, but no Hausdorffness.

  • $\begingroup$ yeah, that was also my proof. $\endgroup$ Jul 4, 2011 at 22:08

To QUESTION 1: Yes, exponential law can be established without Hausdorff property of the middle space.

In fact the problem consists of four different parts each requiring its own arguments. Let me try to clarify the points.

For sets $ M, N $ denote by $Map( M, N )$ the set of all maps $ M\to N $. Let $ L $ be another set. Then there is a bijection $$ \exp:Map( L \times M, N ) \to Map( L , Map( M, N ))$$ given by the following rule: if $ F \in Map( L \times M, N )$, then $\exp( F ): L \to Map( M, N )$ is given by $\exp( F )(t)(x) = F (t,x)$.

Lemma. Let $ L , M, N $ be topological spaces. Then the following statements hold.

(1) $\exp(C(L \times M, N)) \subset C(L,C(M,N))$, i.e. if $F: L \times M \to N$ is continuous, then $\exp(F): L \to C(M,N)$ is continuous as well.

(2) The induced map $$ \exp:C(L \times M, N) \to C(L,C(M,N)) $$ is continuous with respect to the corresponding compact open topologies.

(3) Suppose $ M$ is locally compact in the sense that for each $x\in M$ and an open neighborhood $ U $ of $x$ there exists a compact subset $ B \subset M$ such that $x\in Int{ B } \subset B \subset U $, (notice that it is not required that $ M$ is also Hausdorff). Then $$ \exp (C(L \times M, N)) = C(L,C(M,N)),$$ i.e. the above map is a continuous bijection.

(4) If in addition to (3) $ L $ is a $T_3$-space (again not necessarily Hausdorff) then the map $$ \exp: C(L \times M, N) \to C(L,C(M,N)) $$ is a homeomorphism.

Proof. (1) Let $ F : L \times M \to N $ be a continuous map. We need to prove that $ f = \exp( F ): L \to C(M,N)$ given by $ f (t)(x) = F (t,x)$ is continuous. Let $t\in L $ and $$ [ B , W ] = \{ g \in C(M,N) \mid g( B )\subset W \} $$ be an open neighborhood of $ f (t)$ in $C(M,N)$. We need to find an open neighborhood $ U $ of $t$ in $ L $ such that $ f ( U ) \subset [ B , W ]$

This means that $ f (t)( B ) \subset W $, that is $t\times B \subset F ^{-1}( W )$. Since $ B $ is compact, there exists an open neighborhood $ U $ of $t$ in $ L $ such that $ U \times B \subset F ^{-1}( W )$. But this means that for each $s\in U $, $ f (s)( B )\subset W $, i.e. that $ f ( U ) \subset [ B , W ]$.

(2) Let $ F : L \times M \to N $ be a continuous map and $ f = \exp( F ): L \to C(M,N)$ is given by $ f (t)(x) = F (t,x)$ is continuous. Let also $A \subset L $ and $ B \subset M$ be compact subsets, $ W \subset N $ be an open set, and $$ [A, [ B , W ]] = \{ g: L \to C(M,N) \mid g(A)( B ) \subset W \} $$ be an open neighborhood of $ f $ in $C(L,C(M,N))$. Then $[A\times B , W ]$ is an open neighborhood of $ F $ in $C(L \times M, N)$ such that $$ \exp([A\times B , W ]) \subset [A, [ B , W ]]. $$

(3) Suppose $ M$ is locally compact in the above sense and let $ f : L \to C(M,N)$ be a continuous map. We should prove that $ F = \exp^{-1}( f ): L \times M \to N $, $ F (t,x)= f (t)(x)$ is continuous. Let $(t,x)\in L \times M$ and $ W $ be an open neighborhood of $ F (t,x)$ in $ N $. We need to find neighborhoods $ U \subset L $ and $V \subset M$ of $t$ and $x$ such that $ F ( U \times V)\subset W $.

Since $ f (t): M\to N $ is continuous, $ f (t)^{-1}( W )$ is an open neighborhood of $x$ in $ M$, whence there exists a compact $ B \subset M$ such that $x\in Int{ B } \subset B \subset U $. Then $[ B , W ]$ is an open neighborhood of $ f (t)$, and since $ f : L \to C(M,N)$ is continuous at $t$, there exists an open neighborhood $ U \subset L $ of $t$ such that $ f ( U )\subset [ B , W ]$. This implies that $ F ( U \times B ) \subset W $. In particular, one can put $V= Int{ B }$, then we will have $ F ( U \times V) \subset W $.

(4) Suppose $ L $ is a $T_3$ space, i.e. for each point $x\in L $ and its open neighborhood $ U $ there exists an open neighborhood $ Z$ such that $x\in Z\subset\overline{ Z} \subset U $, and $ M$ is locally compact as in (3). Let $ F \in C(L \times M, N)$, $ f = \exp( F )$, and $[C, W ]$ be an open neighborhood of $ F $, where $C\subset L \times M$ be a compact subset and $ W \subset N $ is open. We need to prove that $\exp([C, W ])$ contains an open neighborhood of $ f $ in $C(L,C(M,N))$.

Since $C$ is compact and $ F ^{-1}( W )$ is its open neighborhood in $ L \times M$, there exist finitely many open sets $ U _1,\ldots, U _k \subset L $, and $V_1,\ldots,V_k \subset M$, such that $$ C \subset \cup_{i=1}^{k} U _i \times V_i \subset F ^{-1}( W ). $$ Moreover, since $ L $ is $T_3$ and $ M$ is locally compact, one can assume in addition that

(a) each $V_i = Int{ B _i}$ for some compact $ B _i \subset M$;

(b) and that $$ C \ \subset\ \cup_{i=1}^{k} U _i \times V_i \ \subset \ \cup_{i=1}^{k} \overline{ U _i} \times B _i \ \subset \ F ^{-1}( W ). $$

Let $p: L \times M\to L $, $p(t,x)=t$, be the natural projection. Then $p(C)$ is compact. Hence $A_i := \overline{ U _i} \cap p(C)$ is a closed subset of $p(C)$ and therefore compact. It then follows that $$ C \subset \cup_{i=1}^{k} A_i \times B _i \subset F ^{-1}( W ). $$ This also implies that $$ \cup_{i=1}^{k} [A_i, [ B _i, W ]] \ \subset \exp( [C, W ] ). $$ This proves that $\exp$ is open and therefore a homeomorphism.

Thus for the proof of (1) and (2) we do not need any assumptions on $L$ and $M$; for (3) we need local compactness of $M$ and for (4) we need an additional $T_3$ property of $L$.


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