Take the 2-minute tour ×
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.

Let $X$ and $Y$ be topological spaces. How to compare $\tau_{C(Y,X)}$ and $\tau_{co}$?

(By definition, $C(Y,X):=\{f:Y\rightarrow X|\textit{$f$ is continuous}\}\subset X^Y$. For $K\subset Y$ and $U\subset X$ is defined $M(K,U):=\{f\in C(Y, X)|f(K)\subset U\}$. Compact-open topology $\tau_{co}$ on $C(Y,X)$ is topology defined by subbasis $S_{co}:=\{M(K, U)|K\in K_Y, U\in\tau_X\}$).

Any help is welcome. Thanks in advance.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Let me write $C$ for $C(X,Y)$ equipped with the compact open topology, and $C'$ for $C(X,Y)$ equipped with subspace topology of the product topology on $Y^X$.

The canonical bijection $$\theta:C\to C',\quad f\mapsto f$$ is continuous, i.e., the topology induced from the product topology is coarser than the compact open topology.

Indeed, $\theta$ is continuous iff $i\circ\theta:C\to C'\to Y^X$ is continuous (where $i:C\to Y^X$ is the canonical injection) iff $p_x\circ i\circ\theta:C\to C'\to Y^X\to Y$ is continuous for all $x\in X$. If $U\subset Y$ is an open subset, then $$\big(p_x\circ i\circ\theta\big)^{-1}(U)=M(\lbrace x\rbrace,U)$$ is a subbasic open subset of $C$ by definition of the compact open topology (singletons are compact.)

share|improve this answer
    
Thanks on swift answer. So, $\tau_{C(X,Y)}\subset\tau_{co}$, but does the opposite hold (are those topologies equivalent or not)? –  alans Dec 24 '13 at 0:10
1  
Generally, the compact open topology is strictly finer. I might be wrong, but the only case when they coincide is when the only compact subsets of $X$ are finite, or $Y$ is trivial. –  Olivier Bégassat Dec 24 '13 at 0:12
    
Is there any counterexample which shows that other inclusion doesn't hold in general case? –  alans Dec 24 '13 at 21:49

Your Answer

 
discard

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.