CW-complex, Hausdorff space, second-countable space, sober space
connected space, locally connected space, contractible space, locally contractible space
hom-set, hom-object, internal hom, exponential object, derived hom-space
loop space object, free loop space object, derived loop space
A natural topology on mapping spaces of continuous functions, important because of its role in exhibiting locally compact topological spaces to be exponentiable, as demonstrated below, culminating in Corollary 1.
The compact-open topology on the set of continuous functions $X \to Y$ is generated by the subbasis of subsets $U^K \subset C(X,Y)$ that map a given compact subspace $K \subset X$ to a given open subset $U \subset Y$, whence the name.
When restricting to continuous functions between compactly generated topological spaces one usually modifies this definition to a subbase of open subsets $U^{\phi(K)}$, where now $\phi(K)$ is the image of a compact topological space under any continuous function $\phi \colon K \to X$. This definition gives a cartesian internal hom in the category of compactly generated topological spaces (see also at convenient category of topological spaces).
The two definitions agree when the domain $X$ is a compactly generated Hausdorff space, but not in general. Beware that texts on compactly generated spaces nevertheless commonly say “compact-open topology” for the second definition.
Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. We denote by $Y^{X}$ the set of continuous maps from $(X,\mathcal{O}_{X})$ to $(Y, \mathcal{O}_{Y})$.
Other common notations for $Y^{X}$ are $Map(X,Y)$ or $C(X,Y)$.
Let $(X, \mathcal{O}_{X})$ be a topological space. We denote by $\mathcal{O}^{c}_{X}$ the set of subsets of $X$ which are compact with respect to $\mathcal{O}_{X}$.
Let $(X, \mathcal{O}_{X})$ be a topological space. Let $A$ be a subset of $X$. We denote by $\overline{A}$ the closure of $A$ in $X$ with respect to $\mathcal{O}_{X}$.
A topological space $(X, \mathcal{O}_{X})$ is locally compact if, for every $x \in X$ and every $U \in \mathcal{O}_{X}$ such that $x \in U$, there is a $V \in \mathcal{O}_{X}$ such that $x \in V$, such that $\overline{V} \in \mathcal{O}^{c}_{X}$, and such that $\overline{V} \subset U$.
There are many variations on this definition, which can be found at locally compact space. These are all equivalent if $(X, \mathcal{O}_{X})$ is Hausdorff. We do not however make the assumption that $(X, \mathcal{O}_{X})$ is Hausdorff.
Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. Given $A \in \mathcal{O}^{c}_{X}$ and $U \in \mathcal{O}_{Y}$, we denote by $M_{A,U}$ the set of continuous maps $f : X \rightarrow Y$ such that $f(A) \subset U$.
Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. The compact-open topology on $Y^{X}$ is that with sub-basis given by the set of sets $M_{A,U}$ such that $A \in \mathcal{O}^{c}_{X}$ and $U \in \mathcal{O}_{Y}$.
Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. We shall denote the compact-open topology (def. 2) on $Y^{X}$ by $\mathcal{O}_{Y^{X}}$.
Let $(X, \mathcal{O}_{X})$ be a locally compact topological space, def. 1, and let $(Y, \mathcal{O}_{Y})$ be a topological space. The map $ev : X \times Y^{X} \rightarrow Y$ given by $(x,f) \mapsto f(x)$ is continuous, where $X \times Y^{X}$ is equipped with the product topology $\mathcal{O}_{X \times Y^{X}}$ with respect to $\mathcal{O}_{X}$ and $\mathcal{O}_{Y^{X}}$ (notation 5).
By an elementary fact concerning continuous maps, it suffices to show that for any $(x,f) \in X \times Y^{X}$, and any $U \in \mathcal{O}_{Y}$ such that $f(x) \in U$, there is a $U' \in \mathcal{O}_{X \times Y^{X}}$ such that $(x,f) \in U'$, and such that $ev(U') \subset U$.
To demonstrate this, we make the following observations.
1) Since $f$ is continuous, we have that $f^{-1}(U) \in \mathcal{O}_{X}$.
2) Since $(X, \mathcal{O}_{X})$ is locally compact, we deduce from 1) that there is a $V \in \mathcal{O}_{X}$ such that $x \in V$, such that $\overline{V} \in \mathcal{O}^{c}_{X}$, and such that $V \subset f^{-1}(U)$.
3) By 2) and by definition of $\mathcal{O}_{X \times Y^{X}}$ and $\mathcal{O}_{Y^{X}}$, we have that $V \times M_{\overline{V}, U} \in \mathcal{O}_{X \times Y^{X}}$, and that $(x,f) \in V \times M_{\overline{V}, U}$.
4) Let $(x',f') \in V \times M_{\overline{V}, U}$, Since $x' \in V$, and since $f'(V) \subset f'(\overline{V}) \subset f(U)$, the latter inclusion holding by definition of $M_{\overline{V}, U}$, we have that $f'(x) \in U$. We deduce that $ev(V \times M_{\overline{V}, U}) \subset U$.
By 3) and 4), we see that we can take $U'$ as the beginning of the proof to be $V \times M_{\overline{V}, U}$.
Let $(X, \mathcal{O}_{X})$, $(Y, \mathcal{O}_{Y})$, and $(Z, \mathcal{O}_{Z})$ be topological spaces. Let $f : X \times Y \rightarrow Z$ be a continuous map, where $X \times Y$ is equipped with the product topology $\mathcal{O}_{X \times Y}$ with respect to $\mathcal{O}_{X}$ and $\mathcal{O}_{Y}$. Then the map $\alpha_{f} : X \rightarrow Z^{Y}$ given by $x \mapsto f_{x}$ is continuous, where $f_{x} : Y \rightarrow Z$ is given by $y \mapsto f(x,y)$, and where $Z^{Y}$ is equipped with the compact-open topology $\mathcal{O}_{Z^{Y}}$.
By an elementary fact concerning continuous maps, it suffices to show that for any $x \in X$, and any $M_{A,U} \in \mathcal{O}_{Z^{Y}}$ such that $f_{x} \in M_{A,U}$, there is a $U' \in \mathcal{O}_{X}$ such that $x \in U'$, and such that $\alpha_{f}( U' ) \subset M_{A,U}$.
To demonstrate this, we make the following observations.
1) Since $f_{x} \in M_{A,U}$, we have, by definition of $f_{x}$ and by definition of $M_{A,U}$, that $f(x,a) \in U$ for all $a \in A$.
2) Since $f$ is continuous, we have that $f^{-1}(U) \in \mathcal{O}_{X \times Y}$.
3) By 1), we have that $\{x\} \times A \subset f^{-1}(U)$.
4) Let $\mathcal{O}_{X \times A}$ denote the subspace topology on $X \times A$ with respect to $\mathcal{O}_{X \times Y}$. By 2), we have that $f^{-1}(U) \cap (X \times A) \in \mathcal{O}_{X \times A}$.
5) By 3), we have that $\{x \} \times A \subset f^{-1}(U) \cap (X \times A)$.
6) Since $A \in \mathcal{O}^{c}_{Y}$, it follows from 4), 5), and the tube lemma that there is a $U' \in \mathcal{O}_{X}$ such that $x \in U'$ and such that $U' \times A \subset f^{-1}(U) \cap (X \times A) \subset f^{-1}(U)$.
7) We deduce from 6) that $f(U' \times A) \subset U$. This is the same as to say that $f_{x'}(a) \in U$ for all $x' \in U'$. Thus $f_{x'}$ belongs to $M_{A,U}$ for all $x' \in U'$, which is the same as to say that $\alpha_{f}(U') \subset M_{A,U}$.
We conclude that we can take the required $U'$ of the beginning of the proof to be the $U'$ of 6).
Let $(X, \mathcal{O}_{X})$, $(Y, \mathcal{O}_{Y})$, and $(Z, \mathcal{O}_{Z})$ be topological spaces. Suppose that $(X, \mathcal{O}_{X})$ is locally compact. Let $f : X \rightarrow Z^{Y}$ be a continuous map, where $Z^{Y}$ is equipped with the compact-open topology $\mathcal{O}_{Z^{Y}}$. Then the map $\alpha^{-1}_{f} : X \times Y \rightarrow Z$ given by $(x,y) \mapsto \big( f(x) \big)(y)$ is continuous, where $X \times Y$ is equipped with the product topology $\mathcal{O}_{X \times Y}$.
We make the following observations.
1) We have that $\alpha^{-1}_{f} = ev \circ \tau \circ (f \times id)$, where $\tau : Z^{Y} \times Y \rightarrow Y \times Z^{Y}$ is given by $(f,y) \mapsto (y,f)$.
2) By Proposition 10, we have that $ev$ is continuous.
3) It is an elementary fact that $\tau$ is continuous.
4) Since $f$ is continuous, it follows by an elementary fact that $f \times id$ is continuous.
5) We deduce from 2) - 4) that $ev \circ \tau \circ (f \times id)$ is continuous. By 1), we conclude that $\alpha^{-1}_{f}$ is continuous.
Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. Suppose that $(X, \mathcal{O}_{X})$ is locally compact. Then $(Y^{X}, \mathcal{O}_{Y^{X}})$ together with the corresponding map $ev$ defines an exponential object in the category $\mathsf{Top}$ of all topological spaces.
Follows immediately from Proposition 2, Proposition 3, and the fact that $Y^{X}$ and $ev$ are exhibited by the corresponding maps $\alpha_{-}$ and $\alpha^{-1}_{-}$ of Proposition 2 and Proposition 3 to define an exponential object in the category $\mathsf{Set}$ of sets.
A proof can also be found in Aguilar-Gitler-Prieto 02, prop. 1.3.1, or just about any half-decent textbook on point-set topology!
However, the result is almost universally stated with an assumption that $(X,\mathcal{O}_{X})$ is Hausdorff which, as the proof we have given illustrates, is not needed.
We moreover have a homeomorphism $Map(X,Map(Y,Z))\cong Map(X\times Y,Z)$ if in addition $X$ is Hausdorff. See also convenient category of topological spaces.
If $Y$ is a metric space then the compact-open topology on $Map(X,Y)$ is the topology of uniform convergence on compact subsets in the sense that $f_n \to f$ in $Map(X,Y)$ with the compact-open topology iff for every compact subset $K\subset X$, $f_n \to f$ uniformly on $K$. If (in addition) the domain $X$ is compact then this is the topology of uniform convergence.
Ralph H. Fox, On Topologies for Function Spaces , Bull. AMS 51 (1945) pp.429-432. (pdf)
Eva Lowen-Colebunders, Günther Richter, An Elementary Approach to Exponential Spaces, Applied Categorical Structures May 2001, Volume 9, Issue 3, pp 303-310 (publisher)
Marcelo Aguilar, Samuel Gitler, Carlos Prieto, sections 1.2, 1.3 of Algebraic topology from a homotopical viewpoint, Springer (2002) (toc pdf)