3

We equip sets of continuous functions $C(X,Y)$ with compact-open topologies, which are topologies generated by prebasis of sets of form $$ W(K,U) = \{ f \in C(X,Y) : f(K) \subset U\}, $$ where $K$ is a compact subset of $X$ and $U$ is an open subset of $Y$.

The problem is to show that spaces with such topology $C(X,Y)\times C(X,Z)$ and $(X,Y \times Z)$ are homeomorphic.

I want to show that map

$$\varphi : C(X,Y)\times C(X,Z) \to C(X,Y \times Z)$$

$$ \varphi : (f,g) \mapsto f \times g $$

$$ \varphi(f,g)(x) := (f(x),g(x)) $$

is a homeomorphism.

From the universl property of product we can infer that $\varphi^{-1}$ exists and continuous.

On the other hand I don't know how to show, that $\varphi$ is continuous. Here is my work so far:

$C(X,Y \times Z)$ has a topology generated by sets of the form $W(K,U)$, where $K$ is compact and $U = \bigcup_{i \in I} V_i \times W_i$, with $V_i$ and $W_i$ being open in $Y$ and $Z$ respectively. If $f \in W(K,U)$, then $f(K)$ is compact as $f$ is continuous, which means that $f(K)$ can be covered by a finite set of rectangles $\; (V_{i_j} \times W_{i_j})^n_{j=1}$. So I want to claim that $f$ has an open neighborhood in $W(K,U)$ of of the form $ \bigcap_{(k,j) \in J }W(C_k, V_{i_j} \times W_{i_j} )$ for some finite set $J$ and a collection of compacts $C$. But I don't know how to construct them.

I noticed that it is possible to construct compact sets

$$ \alpha_k = f^{-1}\left(Y \times Z \setminus \bigcup^n_{j = 1 , j \neq k} V_{i_j} \times W_{i_j}\right) \cap K, $$ But they are insufficient as they cover only points in $x \in K$, such that $f(x)$ does not belong to any intersection of said open rectangles. I don't know how to construct compacts which will correspond to intersections.

I know how to prove this result in case $X$ is Hausdorff, but I heard that it is also true in general.

Nik Bren
  • 1,993
  • 1
  • 19
  • 35
  • 1
    Are you sure that it is true in general? If so then who (or what paper) told you? I am just curious. If $X$ is Hausdorff and $\mathcal V$ is a subbase of $Y\times Z$ (for instance the sets $U\times V$ with $U$ open in $Y$ and $V$ open in $Z$) then it can be shown that sets $W(K,U)$ with $K\subseteq X$ compact and $U\in\mathcal V$ form a subbase of $(X,Y\times Z)$. I suspect though that you allready know that, because you mention to know a proof for that case. – drhab Mar 07 '18 at 13:57
  • 1
    @drhab , honestly, this is the statement from the lecture I was attending, and Professor that unlike exponential law this result does not require any restrictions to any spaces. And he left proof to the listeners. I have been working over this problem almost for two weeks already. The result probably might be false. But then I need to find a counter-example with non-Hausdorff $X$. – Nik Bren Mar 07 '18 at 19:44
  • @NikPronko this fact is not the exponential law. – Henno Brandsma Mar 08 '18 at 23:04
  • @drhab this fact is a proposition in Engelking, even more generally, namely 3.4.5 (left to the reader to show..) – Henno Brandsma Mar 08 '18 at 23:04

1 Answers1

3

To see the continuity of $\varphi$: let $W(K, U \times V)$ be a subbasis element of $C(X, Y \times Z)$, (where $U \subseteq Y$ and $V \subseteq Z$ are open) and suppose $(f,g) \in O:=\varphi^{-1}[W(K,U \times V)]$. We want to show $(f,g)$ is an interior point of $O$.

(Lemma 3.4.6 in Engelking (General Topology, 2nd ed.) says that in a compact-open $C(X,Y)$ we can take all sets $W(K,O)$ as a subbase where we choose the $O$ not all opens but from a fixed subbase for $Y$, this is handy to reduce the number of open sets to consider in proofs.)

Note that $f \in W(K, U)$ and $g \in W(K,V)$ and that the open neighbourhood $W(K,U) \times W(K,V)$ is mapped by $\varphi$ into $W(K, U \times V)$, as is easily seen, so indeed $(f,g) \in W(K,U) \times W(K,V) \subseteq O$, as required.

ChoF
  • 1,740
Henno Brandsma
  • 250,824
  • 1
    This is exactly what I had in mind as proof. But I could only give a proof of the lemma under the extra condition that $X$ is Hausdorff. The declaration of the OP ("I know how to prove this result in case $X$ is Hausdorff...") seemed to confirm the necessity of that condition. It is pleasing to hear that it is not necessary after all. I would like to get hold of a proof of the lemma, but have not the disposal of Engelking. Of course I could ask it as a question on this site, but do you know a more direct route maybe? – drhab Mar 09 '18 at 08:21
  • @drhab Engelking assumes that compact sets like $K$ are Hausdorff (as a general principle) but in this theorem has no extra assumption on $X$. – Henno Brandsma Mar 09 '18 at 10:08
  • So if I understand well then according to Engelking the sets $W(K,U)$ form a subbase for compact-open topology if the $U$ ranges over open subsets and $K$ over - what I would call - compact Hausdorff sets. In that Engelking follows Bourbaki. Is my understanding correct? – drhab Mar 09 '18 at 10:30
  • @drhab you’re correct. – Henno Brandsma Mar 09 '18 at 10:43
  • 1
    @drhab so for a cofinite space, the compact open topology is just the pointwise topology. – Henno Brandsma Mar 09 '18 at 10:45
  • Thank you. Quite recently I encountered (in May's book) the Bourbaki setting with compact vs compact-Hausdorff and quasi-compact vs compact. I dislike the fact that there are $2$ possible interpretations, but must learn to live with it. This encourages me to be very cautious in the future. – drhab Mar 09 '18 at 10:50
  • Thank you for the reference. It seems that this difference in notion of compactness was indeed a source of my confusion. I keep getting flabbergasted by it every time I head on to learn algebraic topology. – Nik Bren Mar 09 '18 at 11:03
  • 1
    @NikPronko local compactness is even more confusing. It can mean (depending on the author) 1. every point has a compact neighbourhood. 2. Every point has an open neighbourhood with compact closure. 3. Every point has a local base of compact neighbourhoods. These coincide for Hausdorff spaces, but not in general. (in the previous: compact is just covering-compact without Hausdorff). That's one of the reasons that locally compact and Hausdorff are often assumed together. – Henno Brandsma Mar 09 '18 at 12:25