Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If the function $\varphi :Z\rightarrow C(X,Y)$ ($C(X,Y)$ with compact-open topology) is continuous and $X$ is locally compact, then $$F\colon Z\times X\rightarrow Y$$ $$F(z,x)=\varphi (z)(x)$$ will be continuous.

My idea was to show that $F^{-1}(U)$ is open when $U$ is open in $Y$.

  1. What is the elements of $F^{-1}(U)$?

  2. I want to take any point in $F^{-1}(U)$ and show that for this point, we can find an open ball contained it, but I don't know how I should do that and also when I must use locally compactness of $X$.

  3. I have no Imagination about this spaces, can you help me with this?

Please help me with your knowledge. This is a very important problem for me. Thank you very much.

share|cite|improve this question
up vote 4 down vote accepted

So you have $φ$ and want to show continuity of $$F:Z×X\to Y\\ (z,x)\mapsto φ(z)(x)$$ The idea is to write $F$ as the composition of two maps, namely $$Z\times X\to C(X,Y)×X\to Y$$ where the first map is $\varphi×1_X$ which is continuous by assumption, and the second map is the evaluation $$ε:C(X,Y)×X\to Y\\ (f,x)\mapsto f(x)$$ Now $ε$ is continuous, and the proof goes as follows:
Take a point $(f,x)$ where $f:X\to Y$ and $x\in X$, and consider an open neighbourhood $V$ of $ε(f,x)=f(x)$. By continuity of $f$, there is an open neighbourhood $U$ of $x$ such that $f[U]⊆V$. By local compactness of $X$, there is also a compact neighbourhood $K$ with $x\in K⊆U$. Then $(K,V)×K$ is a neighbourhood of $(f,x)$ such that $ε(g,y)=g(y)\in V$ for $g\in(K,V)$ and $y\in K$. Hence $ε$ is continuous.

There is a converse to this: Given a continuous $F:Z×X\to Y$, then $$\hat F:Z\to C(X,Y)\\ \hat F(z)(x)=F(z,x)$$ is continuous. To show this, let $z\in Z$ and $(K,V)$ a subbase neighborhood of $\hat F(z)$. Then $\hat F(z)(x)\in V$ for all $x\in K$, thus $\{z\}×K⊆F^{-1}[V]$ which is open. By the Tube Lemma we can find an open $U$ such that $\{z\}×K⊆U×K⊆F^{-1}[V]$. Then $U$ is a neighbourhood of $z$ such that $\hat F[U]⊆(K,V)$.

Furthermore, now that we have this $\hat F$ we can consider $$Z×X\xrightarrow{\hat F×1_X}C(X,Y)×X\xrightarrowεY$$ and we see that this gives us our original $F$, which means that turning $F$ into $\hat F$ is sort of a converse to turning a $φ:Z\to C(X,Y)$ into $F$. After all we get a bijection $$C(Z,C(X,Y))\cong C(Z×X,Y)$$

To get a picture of how a continuous map $\phi:Z\to C(X,Y)$ looks like, you could take as an example the case $X=Y=Z=\Bbb R$. A map $\Bbb R\to\Bbb R$ can be thought of as an unbroken curve without "jumps". A function $\Bbb R\to C(\Bbb R,\Bbb R)$ assigns to each real number $z$ such a curve, let's denote the function assigned to $z$ by $f_z$. We can put these $f_z$ side by side, so that the graph of $f_z$ "hovers" over the line $\{z\}×\Bbb R$. All those maps together give a function $\Bbb R×\Bbb R\to\Bbb R$, namely the function $(z,x)\mapsto f_z(x)$. There is no reason why $F$ should be continuous if $φ$ is merely a function, but the topology on $C(\Bbb R,\Bbb R)$ allows us to demand a continuous $φ$. Intuitively, $φ$ is continuous if for small changes in $z$, the maps $f_z$ do not change drastically. It turns out that the compact-open topology is the suitable one. I invite you to figure out what a continuous $φ$ means if the topology on $C(\Bbb R,\Bbb R)$ is the product topology. Actually, in that case this makes the map $F$ only continuous in both coordinates $z$ and $x$, but not continuous overall.

share|cite|improve this answer
thank you very much,it it great,I really feel that I learned a lot. – kpax Feb 16 '14 at 2:48

The following paper.available here, is relevant:

Booth, P.I. and Tillotson, J. "Monoidal closed categories and convenient categories of topological spaces", Pacific J. Math. 88 (1980) 33--53.

It discusses topologies on the product $X \times Y$ and on the set $C(Y,Z)$ of continuous functions $Y \to Z$ so that one gets a natural bijection $$C(X \times Y,Z) \cong C(X,C(Y,Z)). $$

There is a modification of the compact-open topology to a "test-open" topology in which one considers a class $\mathcal A$ of spaces and defines $C_{\mathcal A}(Y,Z)$ to be the set of functions $f: Y \to Z$ such that $f\circ t: A \to Z$ is continuous for all continuous functions $t: A \to Y $ and all $A \in \mathcal A$. A standard example is to consider $\mathcal A$ as the set of all compact Hausdorff spaces. One then gets a natural bijection $$C_{\mathcal A}(X \times_{\mathcal A} Y,Z) \cong C_{\mathcal A}(X,C_{\mathcal A} (Y,Z))$$ where $X \times _{\mathcal A} Y$ is a modified product topology.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.