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Let $S$ denote a set. Then the vector space $FS$ freely generated by $S$ can be identified with the set of all finitely-supported functions $S \rightarrow \mathbb{R}$. This gave me the following idea; assume $X$ is a topological space. Then perhaps it is worth studying the set $C_0(X)$ consisting of all compactly-supported continuous maps $X \rightarrow \mathbb{R}$. This is a vector space in its own right, because:

  • finite unions of compact sets are compact
  • finite linear combinations of continuous functions are continuous.

Note that if $X$ is a discrete space, then $C_0(X)$ agrees with $FX$, because the compact subsets of a discrete space are precisely the finite subsets. (And also because every function out of a discrete space is continuous.)

Question. Does $C_0(X)$ satisfy an interesting universal property?

One idea is that we should be thinking about the forgetful functor from the category of topological vector spaces to the category of topological spaces. If there's a sensible way of equipping $C_0(X)$ with a topology, then perhaps it is left-adjoint to the forgetful functor.

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  • $\begingroup$ I wonder if $F^{\sharp}X$ is something like the free topological vector space on $X$. Don't have time to work out the details at the moment, but it would be interesting in any case to see what the left-adjoint to the forgetful functor from topological vector spaces to topological spaces looks like. $\endgroup$ – Clive Newstead Mar 23 '15 at 16:11
  • $\begingroup$ Doesn't compact-open topology work? $\endgroup$ – user40276 Apr 1 '15 at 3:51
  • $\begingroup$ pretty nonstandard notation $F^\# X$; this is usually denoted $C_0(X)$ or something like it. $\endgroup$ – Mister Benjamin Dover Apr 1 '15 at 22:50
  • $\begingroup$ @Herrmann, any idea what its used for? I've edited with your suggestion. $\endgroup$ – goblin Apr 2 '15 at 2:37
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    $\begingroup$ @goblin: what $C_0(X)$ is used for? It comes up all over the place in analysis (not sure if I understand your question). In any case, in analysis the functorial point of view is not so common (for better or for worse) $\endgroup$ – Mister Benjamin Dover Apr 4 '15 at 11:41
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This isn't a full answer, just a simple observation that spectral theory gives a partial answer of sorts (as you may already know). See here or any introduction to Banach algebra/C*-algebra theory for supporting details.

A character of a unital commutative Banach algebra $A$ is a unital algebra homomorphism $\phi: A \to \mathbb{C}$. The set $\Phi_A$ of all characters of $A$ has a canonical topology (the relative weak-star topology) which makes it a compact Hausdorff space called the spectrum of $A$. In fact, the assignment $A \mapsto \Phi_A$ is a contravariant functor from unital commutative Banach algebras to compact Hausdorff spaces.

If $X$ is a compact Hausdorff space (one could get away with less, but this is convenient) then $C(X)$ is a unital commutative Banach algebra (in fact a C*-algebra). There is a mapping (actually a homeomorphism) $X \to \Phi_{C(X)}$ sending $x$ to evaluation at $X$. This map $$X \to \Phi_{C(X)}$$ is universal in the sense that, if $A$ is another unital commutative Banach algebra and there is a given (continuous) map $$x \mapsto \phi_x : X \to \Phi_A,$$ then there is a unique homomorphism $a \mapsto f_a: A \to C(X)$ which makes the diagram $$\begin{matrix} X & \to & \Phi_A \\ & \searrow & \uparrow \\ & & \Phi_{C(X)} \end{matrix}$$ commutative. It is easy enough to see the uniqueness. Commutativity of the diagram entails that, for each $a \in A, x \in X$, we have $f_a(x) = \phi_x(a)$. It is simple to check that $a \mapsto f_a$ is a homomorphism.

I guess my question to you is whether you want a notion of univerality which looks something like this in general? Or if are looking for something wildly different.

Added: Responding to your last comment, yes the universal property of the mapping $X \to \Phi_{C(X)}$ determines $C(X)$ uniquely as a Banach algebra. What follows is just the usual argument for uniqueness of universal objects

  • Suppose that there are two Banach algebras $A_1$ and $A_2$ and given mappings \begin{align*} X \to \Phi_{A_1} && X \to \Phi_{A_2} \end{align*} which are universal in the sense outlined above. We shall show there are unique inverse isomorphisms between $A_1$ and $A_2$ which make the relevant triangles commute.
  • Using the universality of each map in turn, we see there are unique homomorphisms \begin{align*} f : A_1 \to A_2 && g : A_2 \to A_1 \end{align*} which make diagrams \begin{align*} \begin{matrix} X & \to & \Phi_{A_1} \\ & \searrow & \uparrow \\ & & \Phi_{A_2} \end{matrix} && \begin{matrix} X & \to & \Phi_{A_2} \\ & \searrow & \downarrow \\ & & \Phi_{A_1} \end{matrix} \end{align*} commutative (here the vertical maps are induced contravariantly from $f_1$ and $f_2$).
  • Lastly, one notes that the identity map $\mathrm{id}_{A_1}$ and $f \circ g$ are both mappings $A_1 \to A_1$ which make the diagram
    $$ \begin{matrix} X & \to & \Phi_{A_1} \\ & \searrow & \downarrow \\ & & \Phi_{A_1} \end{matrix} $$ commute so that $f \circ g = \mathrm{id}_{A_1}$ by uniqueness and, similarly, that $g \circ f = \mathrm{id}_{A_2}$. Thus, $f$ and $g$ are inverse isomorphisms between the Banach algebras $A_1$ and $A_2$.
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  • $\begingroup$ Interesting! Thanks for the info. $\endgroup$ – goblin Apr 2 '15 at 9:24
  • $\begingroup$ What I don't understand is whether this actually characterizes $C(X).$ It looks like this characterizes the "dual space" $\Phi_{C(X)}.$ If so, then this isn't quite what I was looking for. It seems a bit like giving a universal property for the vector space $\prod_{i \in I}V^i$ when the space of interest is $\coprod_{i \in I} V^i$. $\endgroup$ – goblin Apr 2 '15 at 16:41
  • $\begingroup$ I don't quite understand your comment, but I added an edit to my answer which might clear things up? $\endgroup$ – Mike F Apr 2 '15 at 18:22
  • $\begingroup$ I mean, I don't really get it yet, but that seems to be more an issue with me than with the answer. Consider this accepted. $\endgroup$ – goblin Apr 6 '15 at 12:11
  • $\begingroup$ @goblin: Thanks, although I don't claim this to be a full answer to your question. I was just pointing out a situation where $C(X)$ is universal in some sense. Note this used the ring structure of $C(X)$ where you only asked about the linear (and normed?) structure. Also, of course, you asked about arbitrary topological spaces and used the space of compactly-supported functions. $\endgroup$ – Mike F Apr 6 '15 at 16:23

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