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This question begins with a sort of mysterious comment at the bottom of this Wikipedia page on injective cogenerators. There, it is said, without citation or proof, that as a result of the Tietze Extension Theorem, the interval $I=[0,1]$ is an injective cogenerator for categories of topological spaces satisfying separation axioms (e.g. Hausdorff, Tychonoff, Kolmogorov, or other of the various $T_{i}$ conditions).

So there are a few questions here:

(0) This question is basically terminological. The Wikipedia page says that an injective cogenerator is simply an object that admits a non-zero (although in a general category I'm not sure what a zero map would be anyway) map from every non-zero object. But the nlab page indicates that we should rather define a cogenerator to be an object whose represented functor is faithful, so that's the definition I'm working with for now. Are these equivalent in the case that "zero map" has a meaning?

(1) Is it true that the interval is an injective cogenerator for any of the categories of spaces subject to separation axioms? If so, can you give an idea of how the proof would go? It seems likely that it would work for Tychonoff spaces at least?

(2) If not the interval, do the other categories of spaces subject to separation axioms admit any (set of) cogenerators?

(3) Does the full category of topological spaces have a cogenerator? I was thinking that it might be the two point space $X=\{a,b\}$ with topology given by $\{\{a,b\},\{b\},\emptyset\}$, but I might be wrong here.

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  • $\begingroup$ Your two-point space $X$ is the Sierpinski space, and functions to it are in 1-1 correspondence with open sets in the domain. In fact $Top(-,X)$ represents the functor $Top \to Locale$, and this functor fails to be an equivalence at the spaces that aren't sober (eg non-$T_0$ spaces). $\endgroup$
    – David Roberts ♦
    Commented May 24, 2018 at 2:43
  • $\begingroup$ @DavidRoberts ah ok, so in other words, the Sierpinski space cannot be a cogenerator for all of $Top$? $\endgroup$
    – Jonathan Beardsley
    Commented May 24, 2018 at 2:46
  • $\begingroup$ Oh wow there's a lot written about this Sierpinski space, haha. Okay... so I think I want to say it's a cogenerator of $T_0$-spaces? $\endgroup$
    – Jonathan Beardsley
    Commented May 24, 2018 at 2:47
  • $\begingroup$ Possibly for sober spaces. There are non-sober $T_0$ spaces, I'm pretty sure. $\endgroup$
    – David Roberts ♦
    Commented May 24, 2018 at 3:24
  • $\begingroup$ Also, if you need to use the Tietze extension thm, then presumably the "category of topological spaces subject to separation axioms" should be that of normal spaces. $\endgroup$
    – David Roberts ♦
    Commented May 24, 2018 at 3:31

3 Answers 3

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Perhaps I should have been a little better at Googling before posting this question, but it seems to be answered, to a degree, in a paper from 1980 by Giuli, cited below. In particular, any epi-reflective subcategory, i.e. one that is closed under products and subspaces (hence whose inclusion is limit preserving), has a system of cogenerators that do not form a set. This property is referred to by Giuli as being "weakly initial." This, if I'm reading this right, looks like a system of spaces $X(\Lambda)$ for every cardinal $\Lambda$. Now, in some nice cases, e.g. $T_0$, $T_4$, Tychonoff, and compact Hausdorff, there's in fact a single cogenerator. But this is not the case (according to that paper, though not proven) for $T_1$, $T_2$, $T_3$ or Urysohn spaces. In fact, Giuli says that there are not even systems of cogenerators known for $T_2$, $T_3$, and Urysohn (or $T_{2\frac{1}{2}})$.

However, Giuli does confirm that the interval $[0,1]$ is a cogenerator for Tychonoff spaces and that the Sierpinski space is a cogenerator for $T_0$-spaces (apparently). All of this discussion begins around Theorem 1.2 of that paper.

Giuli, Eraldo, Bases of topological epi-reflections, Topology Appl. 11, 265-273 (1980). ZBL0441.18012.

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    $\begingroup$ Ah, cool. Were you planning to add this to the WP page? $\endgroup$
    – David Roberts ♦
    Commented May 24, 2018 at 7:18
  • $\begingroup$ @DavidRoberts haha I guess I should, to be a good community member and all. $\endgroup$
    – Jonathan Beardsley
    Commented May 24, 2018 at 7:19
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Concerning your third question, the cogenerators of the category of general topological spaces are precisely the non-$T_0$-spaces. See Example 7.18 Remark (4) in Adamek, Herrlich and Strecker's Abstract and concrete categories: The joy of cats (pdf).

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    $\begingroup$ Ah thanks! So it seems then that ANY single non-$T_0$-space, according to that reference, is a cogenerator for $Top$? $\endgroup$
    – Jonathan Beardsley
    Commented May 24, 2018 at 3:43
  • $\begingroup$ @JonathanBeardsley It seems so. $\endgroup$
    – Philippe Gaucher
    Commented May 24, 2018 at 3:55
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    $\begingroup$ @JonathanBeardsley: I haven't looked at the paper, but it's easy to see that this is true, as follows. Let $X$ be your non-$T_0$-space, so that there are $x_1,x_2\in X$ that are topologically indistinguishable. Now for any other space $Y$ and any two different $y_1,y_2\in Y$, you can distinguish them by mapping to $\{x_1,x_2\}\subseteq X$: since continuity is trivial, any bipartition of $Y$ defines such a map, and in particular there is one that distinguishes $x_1$ and $x_2$. This straightforwardly implies that $X$ is a cogenerator. (As the nLab says, coseparator would be a better term.) $\endgroup$
    – Tobias Fritz
    Commented May 24, 2018 at 6:51
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    $\begingroup$ Another way to say this: check that the Sierpinski space is a cogenerator (easy -- it represents open subsets) and observe that the Sierpinski space is a retract of any non-$T_0$ space. $\endgroup$
    – Tim Campion
    Commented May 24, 2018 at 13:35
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    $\begingroup$ Correction to my comment: The indiscrete space $I(2)$ on 2 elements is a cogenerator in $Top$, and that any non-$T_0$ space admits a monomorphism from $I(2)$ so is too. Conversely, no $T_0$ space is a cogenerator with respect to $I(2)$. In $Top_{T_0}$, though, $S$ is a cogenerator, and any $X \in Top_{T_0} \setminus Top_{T_0+T_1}$ has $S$ as a retract so is too. Conversely, no $X \in Top_{T_0+T_1}$ is a cogenerator with respect to $S$. And these turn out to be basically the arguments in The Joy of Cats. $\endgroup$
    – Tim Campion
    Commented Aug 5, 2018 at 14:31
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The answer to your question (0) is "no". The Wikipedia page is in the context of a category with zero object, in which case a zero morphism is one that factors through the zero object. Consider a category with four objects $A,B,C,0$, with $0$ the zero object, and four nonzero nonidentity morphisms $f,g:A\rightrightarrows B$, $h:B\to C$, and $k:A\to C$, where $h\circ f = h\circ g = k$. Then $C$ is a cogenerator in the Wikipedia sense, but $\hom(-,C)$ is not faithful.

The Wikipedia page seems to believe that the two definitions are equivalent at least in an abelian category, since the subsequent sections use instead the property (equivalent to the usual "faithful represented functor" definition) that every object injects into a product of copies of the cogenerator. I'd be surprised if this were true; I would only expect defining (co)generators to detect triviality of objects rather than morphisms to work in contexts like a triangulated category, where whether a morphism is an isomorphism can be detected by whether its cone is zero. The most I can prove from the Wikipedia definition in general is that $\hom(-,C)$ reflects epimorphisms.

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