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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.

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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.)

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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
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

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