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.

  • $\begingroup$ What are the morphisms in $CRing$? $\endgroup$ – Daniel Robert-Nicoud Aug 29 '17 at 10:51
  • $\begingroup$ @DanielRobert-Nicoud, morphisms in $CRing$ are ring structure-preserving homomorphisms. $\endgroup$ – hemidactylus Aug 29 '17 at 10:53
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ I had a feeling that the Stone-Cech compactification involves $C_b(X)$, the ring of bounded continuous functions to $\mathbb{R}$. $\endgroup$ – orangeskid Oct 4 '17 at 12:10
  • $\begingroup$ @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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.