Motivating the compact-open topology It has been a while since I studied algebraic topology, and I wanted to revisit homotopy theory. Determined to take a more sustainable approach, I started by questioning and verifying every result in one of my books, Switzer's Algebraic Topology - Homology and Homotopy.
So in the preliminaries, the compact-open topology on the set $Y^X$ of functions $f: X \to Y$ is defined to be generated by:
$$N_{U,K} = \{f: X \to Y \mid f(K) \subseteq U\},\quad U \subseteq Y \text{ open}, K \subseteq X \text{ compact}$$
One of the principal properties of a topology on $Y^X$ would be that it makes the evaluation mapping $e: Y^X \times X \to Y, e(f,x) = f(x)$ continuous (I know that this applies to the COT only under extra conditions, notably if $X$ is locally compact).
So, for $U \subseteq Y$ open we expect $e^{-1}(U)$ to be open in the product topology. This amounts to, for every $x$ with $f(x) \in U$ for some $f$, the existence of a neighborhood $V_x$ such that there is a neighborhood $T$ in $Y^X$ with $f(V_x) \subseteq U$ for each $f \in T$.
However, taking such $T$s as generators -- explicitly:
$$T_{U,V} = \{f \mid f(V) \subseteq U\} \quad V \subseteq X, U \subseteq Y \text{ open}$$
gives rise to a different topology than the compact-open topology. So what compelling reasons are there to consider the compact-open topology rather than the one I just described? If applicable, historical references are also appreciated.
In particular, I'm interested in results where we can see that the properties of the COT are really "needed" for the proof to follow through.
 A: For starters, the compact-open topology on $\mathscr C(X,Y)$ gives you uniform convergence on compact subsets when $Y$ is metric. The product topology on $Y^X$ gives you pointwise convergence. :)
A: In Theorem 46.11 of Munkres' book on Topology the following is shown. Suppose we take the compact-open topology on $\mathcal C(X,Y)$. If $Z$ is any space and if $f:X\times Z\to Y$ is continuous, then the map $F:Z\to \mathcal C(X,Y)$ is continuous where $F(z)$ is the map given by $x\mapsto f(x,z)$. We say that the map $F$ is induced by $f$.
Let $\mathcal F$ denote the set $\mathcal C(X,Y)$ with some topology $\tau$ which has the property that the evaluation map $e:X\times \mathcal F\to Y$ is continuous where $e(x,f)=f(x)$. By the above theorem we have that the map $E$ which is induced by $e$ is continuous. As $E:\mathcal F\to \mathcal C(X,Y)$ is given by $E(f)(x)=e(x,f)=f(x)$, it follows that $E(f)=f$ and so $E$ is the identity map. Thus, we see that $\tau$ contains the compact-open topology.
We have shown that if a topology $\tau$ on $\mathcal C(X,Y)$ makes the evaluation map  continuous, then it must contain the compact-open topology. In Theorem 46.10 of Munkres' book, is shown that if $X$ is locally compact, the compact-open topology on $\mathcal C(X,Y)$ makes the evaluation map continuous. Thus, it follows that when $X$ is locally compact, the compact-open topology is the smallest topology on $\mathcal C(X,Y)$ which makes the evaluation map continuous.
So if $X$ is not locally compact, it is possible that there is a strictly finer/larger topology on $\mathcal C(X,Y)$ which makes the evaluation map continuous.
