# Is this functor full? Top^op -> CRing (ring of real-valued continuous functions on a topological space)

I am trying to teach myself category theory through Leinster's book ("Basic Category theory", arXiv:1612.09375) and I am currently stuck on a part of Exercise 1.2.28. I cannot tell if a particular (very standard, I reckon) contravariant functor, encountered previously (Example 1.2.11), is full or not. Below I define the functor and then show what I did with it.

For any topological space $X$, let $C(X)$ be the ring of continuous real-valued functions on $X$. For each map f: $X \to Y$, a map $C(f): C(Y) \to C(X)$ is induced, sending each $q \in C(Y): Y \to \mathbb{R}$ to $C(f)(q): X \to \mathbb{R}$ according to composition: $C(f)(q) = q \cdot f$.

So $C$ is a contravariant functor. I was able to establish that it is faithful -- it suffices to consider, for each pair of maps $f$ and $f'$ differing in their values $y=f(x),y'=f'(x) \in Y$ at a point $x \in X$, the images under $C(f)$ and $C(f')$ of any element $q \in C(Y)$ such that $q(y)\neq q(y')$.

When it comes to dis/proving fullness of $C$, however, I am stuck. With a similar construction, that of the $Hom(-,W): Vect_k^{op} \to Vect_k$ contravariant functor (Example 1.2.12), I could find counterexamples proving it is not a full functor, while in the case of $C$ I suspect there might be more constraints, given the richer structure of rings, on the allowed maps between objects $C(Y)$ and $C(X)$, which makes me wonder whether in this case it is true that all such maps are indeed obtained as $C(f)$ for some $f$.

So, is the functor $C$ a full contravariant functor? If so, how can I leverage the underlying ring structure and prove it? If not, is there a simple counterexample that disproves $C$'s fullness?

As far as I could see, the functor $C$ is of great importance in some areas of algebraic topology (I found references to Zariski topology and Hilbert's Nullstellensatz while trying to look for an answer online).

Thanks a lot to everyone helping me in this (probably very trivial) question.

• What are the morphisms in $CRing$? – Daniel Robert-Nicoud Aug 29 '17 at 10:51
• @DanielRobert-Nicoud, morphisms in $CRing$ are ring structure-preserving homomorphisms. – hemidactylus Aug 29 '17 at 10:53
• Then I think it might make a difference if you consider $Top$ as enriched over itself or not. If $C(X),C(Y)$ are topological spaces (e.g. with the compact-open topology), then $C(f)$ will be continuous, and I think it shouldn't be too hard finding a morphism in $CRing$ which is not continuous. Thus, I believe that the functor is not full if $Top$ is seen as a non-enriched category. If you see it as enriched, then I don't know. – Daniel Robert-Nicoud Aug 29 '17 at 10:56

This functor is neither full nor faithful. Your argument that it is faithful presupposes the existence of a function $q$ which separates points in $Y$, which you did not prove exists, and which does not exist in general (note that this implies $Y$ is Hausdorff, at the very least). For a silly class of counterexamples, consider sets equipped with the indiscrete topology, with respect to which every continuous real-valued function is constant. But IIRC there are even Hausdorff counterexamples.

Morphisms $C(Y) \to C(X)$ turn out to correspond to continuous maps $\beta X \to \beta Y$ between the Stone-Cech compactifications of $X$ and $Y$; this is closely related to the commutative Gelfand-Naimark theorem. There are many such maps which are not induced from continuous maps $X \to Y$, for example any constant map taking values in $\beta Y \setminus Y$.

The result above implies that this functor is fully faithful when restricted to compact Hausdorff spaces, which are precisely the spaces for which the natural map $X \to \beta X$ is a homeomorphism.

• I had a feeling that the Stone-Cech compactification involves $C_b(X)$, the ring of bounded continuous functions to $\mathbb{R}$. – orangeskid Oct 4 '17 at 12:10
• @orangeskid: hmm, yeah, I might have been too hasty. It is still true that a morphism $C(Y) \to C(X)$ naturally induces a morphism $C_b(Y) \to C_b(X)$, by restricting to the elements with bounded spectrum (which is a purely algebraic concept and is preserved by algebra homomorphisms), but I don't know off the top of my head whether this is a bijection. – Qiaochu Yuan Oct 5 '17 at 7:42

While not true in general, there are large classes of topological spaces for which the map $$Hom (X,Y) \to Hom( C(Y), C(X))$$ is bijective.

You can try to prove these facts:

1. If $X$ is compact Hausdorff then every maximal ideal of $C(X)$ is of the form $\{f \ | \ f(x) = 0\}$ for some unique $x \in X$.

2. While 1. is not true for a space $X$ like $\mathbb{R}$, it's a good exercise to show that every morphism of rings $C(\mathbb{R})\to \mathbb{R}$ is given by the evaluation at a point. ( this is true for a large family of spaces but show it for $\mathbb{R}$ first).

(In the process, you might have to use the fact that every ring morphism from $\mathbb{R}$ to $\mathbb{R}$ is the identity, the case of $1$-point spaces.