Let $S$ and $S'$ be two topological spaces. It is known that two continuous mappings $f,g:S\to S'$ are homotopic if there is a continous mapping $$F:[0,1]\times S\to S'$$ with $$F(0,\cdot)=f(\cdot)$$ $$F(1,\cdot)=g(\cdot)$$ The continuity of $F$ is defined because we can equip $[0,1]\times S$ with the product topology. But if we look at this another way, I believe it is natural and reasonable to say ($C(S;S')$ is the set of all continuous mappings $S\to S'$) $$\tilde F:[0,1]\to C(S;S')\\ \quad\quad\ t\mapsto F(t,\cdot)$$ defines a continuous family of continuous mappings $S\to S'$. This means $C(S;S')$ should be equipped with a topology. My question is: how can we characterize this topology using the topology on $S$ and $S'$ only, without resorting to the product topology above? Let me rephrase in a few different ways:
(1) How can we characterize the topology on $C(S;S')$?
(2) What is a neighborhood of a mapping $f\in C(S;S')$?
(3) When can we say a sequence $f_n\in C(S;S')$ converges to $f\in C(S;S')$?
I think this is easier if $S$ or $S'$ has additional strcutures. For example when $S'$ is a normed vector space we can define a norm on $C(S;S')$ by $$||F||=\sup_{x\in S}||F(x)||$$ But what about the most general case where $S,S'$ are nothing but topological spaces?
You can topologize $C(S, S')$ in a convenient way by taking the subsets $\mathcal{K}(C,V) = \{f \in C(S, S'): f[C] \subseteq V\} \subseteq C(S,S')$, where $C \subseteq X$ is a compact subspace and $V \subseteq Y$ is open, as a subbasis. The subsequently generated topology is the so-called compact-open topology.