The evaluation map $e : C(X,Y) \times X \to Y$, $e(f,x) = f(x)$

Let $(X,d)$ be a metric space carrying its metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), let $(Y,\mathcal{T}_Y)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and let $C(X,Y)$ carry the compact-open topology (The compact-open topology on $C(X,Y)$ for a metric domain $X$, with subbasis $S(K,V) = \{f : f[K] \subseteq V\}$). The evaluation map is

$e : C(X,Y) \times X \longrightarrow Y, \qquad e(f,x) := f(x),$

the domain carrying the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space) of the compact-open topology on $C(X,Y)$ and the metric topology on $X$.

This is a function. For $f \in C(X,Y)$ and $x \in X$ the value $f(x)$ is a well-determined element of $Y$, and a pair $(f,x)$ of the product determines both entries (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space), so $e$ is defined on all of $C(X,Y) \times X$ with no further condition.

Which topology is meant is part of the definition. Continuity of $e$ is a statement about the pair of topologies on the source and the topology on the target (Continuity of a map of topological spaces at a point and globally), and $C(X,Y)$ carries several topologies on this page. Unless another is named, the topology on $C(X,Y)$ inside an evaluation map is the compact-open one; where a subspace of $C(X,Y)$ is evaluated, it carries the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Separate continuity is immediate; joint continuity is not. For fixed $f \in C(X,Y)$ the map $x \mapsto e(f,x) = f(x)$ is continuous, being $f$ itself. For fixed $x \in X$ the map $f \mapsto e(f,x) = f(x)$ is continuous as well, since for open $V \subseteq Y$ its preimage is $S(\{x\},V)$, a subbasic open set of the compact-open topology, $\{x\}$ being compact (The compact-open topology on $C(X,Y)$ for a metric domain $X$, with subbasis $S(K,V) = \{f : f[K] \subseteq V\}$). What is at issue on this page is joint continuity, that is continuity of $e$ on the product, and that genuinely needs a hypothesis on $X$: it holds when $X$ is locally compact, and this page records as a false statement that it holds for every metric $X$, with an explicit witness.

• The evaluation map is what makes the compact-open topology the right one. Among topologies on $C(X,Y)$ making evaluation continuous, coarser is better and the compact-open topology is the standard choice; its subbasic sets $S(K,V)$ are exactly the conditions the argument for joint continuity consumes, a compact neighbourhood of the point being mapped into the target open set.

• The order of the factors is a convention. Writing $e(f,x)$ rather than $e(x,f)$ costs nothing: the two products are different sets, but the two maps are continuous or not together, since swapping the factors is a homeomorphism by the characteristic property of the product (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice). This page always writes the function first.