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This came up today while thinking about topoi in seminar, as the title suggests my question is;

Is there a finitely complete category with terminal object but NO subobject classifier?

Hopefully if the answer is yes, you can give an interesting example that is useful in some way. One of my professors has a saying that

for every definition you should have an interesting example and an interesting non-example

So I am trying to stick to this.


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Nitpick: a finitely complete category automatically has a terminal object. – Qiaochu Yuan Feb 9 '10 at 22:02
@Qiaochu ugh, your totally right. – B. Bischof Feb 9 '10 at 22:18
Thanks guys, these examples were a lot more obvious that I thought. I probably should have thought a bit longer on this question :/ – B. Bischof Feb 9 '10 at 22:39

You should basically never expect anything to have a subobject classifier -- it's a ridiculously strong condition. Some examples showing how quickly it fails to exist:

  • The category of groups is complete. If this category had a subobject classifier Ω, every subgroup would be the kernel of a map to Ω. But not every subgroup is normal.

  • The category of Hausdorff spaces is complete. If it had a subobject classifier Ω, every subspace would be the preimage of a point in Ω. But not every subspace is closed.

  • The category of rings with identity is complete. The terminal object is the zero ring (with 0 = 1). But the zero ring has no maps to any nonzero ring.

  • The category of compact Hausdorff spaces is complete. As above every closed subspace should be the preimage of a point $\ast$ in Ω under a unique map to Ω. For the subspace {1} in {1,2} to be classified by a unique map, $\Omega\setminus \ast$ must be a single point, so Ω is discrete on two points. But then there is only one map from $S^1$ to Ω, while it has uncountably many subobjects.

Edit: the opposite of any cocomplete category will (with probability 1) give examples as well.

  • The category of groups is cocomplete. A subobject classifier in the opposite category would be a group Ω so that every surjection G -> H is a pushout of Ω -> 1 and Ω -> G. This is the amalgamated free product of G with the trivial group over Ω, that is the quotient of G by the image of Ω. But if Ω is a group surjecting to the kernel of every map, that surjection cannot be unique (just consider Z -> 1).

  • The category of sets (also Hausdorff spaces) is cocomplete. In the opposite category Ω would be a set (space) so that every surjection X -> Y was the quotient of X by the image of a map Ω -> X; this would imply that at most one fiber of X -> Y is not just a single point.

  • The category of commutative rings with identity is cocomplete. A subobject classifier for the opposite category (the category of affine schemes) would be an affine scheme Ω with a $\mathbb{Z}$-point Spec $\mathbb{Z}$ -> Ω so that any monomorphism Y -> X is the base change to $\mathbb{Z}$ of a unique map X -> Ω. This is ridiculous; note that Spec $\mathbb{Q}$ -> Spec $\mathbb{Z}$ is injective, and you'll never get Spec $\mathbb{Q}$ as the fiber product of two $\mathbb{Z}$-points.

  • The partial order category of ordinals is cocomplete. The initial object is 0; but no nonzero ordinal maps to (is ≤) 0. Thus Ω would have to be 0, but then we would have for any α ≥ β that α was the coproduct (supremum) of β with 0, which is false whenever α > β.

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I assume "finitely complete" means "has finite limits". If so, it seems to me that counterexamples are pretty common. Mine would be the category of abelian groups.

A subobject classifier is a monomorphism $u:1\to \Omega$, where $1$ is the terminal object, such that for any subobject $S\to X$, there is a unique map $f_S:X\to \Omega$ such that $S=f^{-1}(1)$.

Notice that this means that any automorphism $f:\Omega \to \Omega$ such that $f\circ u = u$ must be the identity map. So $\Omega$ has no nonidentity automorphisms compatible with $u$.

In the category of abelian groups, the terminal object is also the initial object. So $\Omega$ would be an abelian group with no non-identity automorphisms. Since multiplication by $-1$ is always an automorphism, such a group must be a $Z/2$-vector space. Most of those have non-trivial automorphisms, and the ones which don't are clearly not subobject classifiers.

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Yes, there are many, many finitely complete categories with no subobject classifier. Indeed, I daresay most finitely complete categories lack subobject classifiers (for some notion of "most").

For example, any non-trivial meet semilattice is one (every object admits a morphism into the subobject classifier classifying its identity subobject; however, in a preorder, morphisms are unique. Accordingly, in a preorder with a subobject classifier, every object has a unique subobject, and thus, if there is a terminal object, every object is isomorphic to it).

(I second Yuan's nitpick; a finitely complete category automatically has a terminal object)

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There are also cases where there is no subobject classifier, but there is a weak subobject classifier -- meaning it only classifies monomorphisms with a certain property. Various categories of smooth spaces, as well as simplicial complexes are examples.

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I am pretty sure I proved that the category of groups has no subobject classifier at some point. I will try and edit this post with the proof when I have more time, but I think this is an example for you to think about.

EDIT: Ya, this isn't too bad. If there was a subobject classifier $\Omega$ in Groups, then by looking at the characteristic map of the injection of $0 \rightarrow A$ for each map, you will see by writing out the diagrams that A will have to inject into $\Omega$. But then $\Omega$ is bigger than every cardinal since there is a group of every cardinality. That doesn't fly.

EDIT of EDIT: I guess I should flesh this out in greater detail.

The first thing to note is that the category of groups has a zero object (it's terminal object is also initial). I will write 0 for this.

Say Groups had a subobject classifier $0 \stackrel{true}\longrightarrow \Omega$. Note that the map true must be the unique map out of 0. Also note that the unique map $0 \stackrel{!}\rightarrow A$ is a monomorphism (i.e. injection) for every A. Thus

                                       |      |
                                     ! |      | true
                                       |      | 
                                       \/     \/
                                       A ---->Ω 

is a pullback square, where the lower map, $\chi$, is the characteristic map of !. I claim that $\chi$ is a monomorphism. This is because the ker($\chi$) maps to both A and 0 to make the diagram commute, so the inclusion of ker($\chi$) into A factors through 0 by the definition of a pullback. In other words, the kernel is trivial, so $\chi$ is an injection. Thus every group A admits an injection to $\Omega$ which is bad for set theoretic reasons.

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I don't understand your argument ... could you explain where it would break down if I tried to apply it to Sets? – Tom Church Feb 9 '10 at 23:02
I don't understand either. Why does A have to inject into Omega? (Now would also be a fine time for me to figure out how to write math here, presumably a pretty basic skill... I assume there's some way to enter TeX?) – Sridhar Ramesh Feb 10 '10 at 0:31
Ah, dollar signs. Anyway, what property of the injection $0 \rightarrow A$ tells us that its characteristic map $A \rightarrow \Omega$ must be an injection? – Sridhar Ramesh Feb 10 '10 at 0:46
Does my new update help? – Steven Gubkin Feb 10 '10 at 13:10
That makes perfect sense -- thanks for explaining! – Tom Church Feb 10 '10 at 16:00

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