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I have a question out of my curiosity essentially about the intuition of sub-object classifier.

Sub-objects of object $A$ is defined to be one to one correspondence with morphisms $A \rightarrow \Omega$ for some special $\Omega$ in category $\mathcal{C}$. In set this subobject classifier is easily checked to be $\{0,1\}$.

I have a question though. I have some ideas based on discussions though would like to know more about this. How come it is outgoing information from $A$ to $\{0,1\}$? We know that category is same information as its opposite category. So shouldn't we able to define also sub-objects classifiers as $\Omega \rightarrow A$ as well?

Though this is definitely not the case in $\mathbf{Set}$. I would like to know more why is this not the case.

Shaun
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Karim Mansour
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2 Answers2

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Categories are not always similar to their opposites. For example, the category of nonempty sets has a terminal object but no initial object, whereas its opposite has an initial object but no terminal object.

More relevantly to the current question, just because $\mathcal{C}$ is a topos does not mean that $\mathcal{C}^{op}$ is a topos; indeed, as Andreas says below this only happens if they are trivial. And similarly there is no reason to believe that the subobject classifier condition should be rewriteable in terms of maps into $A$.

Admittedly there is a sense in which $\mathcal{C}$ and $\mathcal{C}^{op}$ are "equally informative:" in the sense of model theory, they have the same definable sets and relations (I'm ignoring size issues here for simplicity). But this is a rather loose notion, which in no way implies having the same basic properties.

Noah Schweber
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  • I see. Thank you for this complete answer. It sheds light on my question. – Karim Mansour Apr 25 '21 at 17:31
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    The only time that both a category $\mathcal C$ and its dual are topoi is when they're trivial ($0\cong1$), and then they actually do have the same $\Omega$. (Proof: In any topos, all morphisms to $0$ are iso, so in our $\mathcal C$, all morphisms out of $1$ are iso. That includes the morphism true$:1\to\Omega$.) – Andreas Blass Apr 25 '21 at 18:49
  • @AndreasBlass Good point, fixed! – Noah Schweber Apr 25 '21 at 18:50
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Short version:

The definition of the subobject classifier can be succinctly stated as "$\Omega$ represents the subobject functor". Since the subobject functor is contravariant, the universal property of $\Omega$ must be given in terms of maps into $\Omega$.

Long version:

For any (well-powered, for size reasons) category $C$, and any object $X$ in $C$, write $\mathrm{Sub}(X)$ for the set of subobjects of $X$ (the monic arrows into $X$, up to isomorphism over $X$). We can turn $\text{Sub}$ into a functor by defining its action on morphisms. Here the most natural choice is to make $\text{Sub}$ contravariant, i.e. into a functor $C^{\text{op}}\to \mathsf{Set}$. Why? Well, given an arrow $f\colon X\to Y$ and a subobject $i\colon S\hookrightarrow Y$, we can pull back $i$ along $f$ to get a subobject $S\times_Y X\hookrightarrow X$. In the category $\mathsf{Set}$, this is exactly the preimage operation.

Now $C$ has a subobject classifier exactly when the functor $\text{Sub}$ is representable. This means there is an object $\Omega$ and a natural isomorphism $\text{Sub}(-)\cong \text{Hom}(-,\Omega)$. The fact that $\text{Sub}$ is a contravariant functor on $C$ means that we have to use the contravariant Hom functor $\text{Hom}(-,\Omega)$ (instead of the covariant Hom functor $\text{Hom}(\Omega,-)$), and thus subobjects of $X$ are classified by arrows $X\to \Omega$.

At this point you might ask if we could turn $\text{Sub}$ into a covariant functor instead. After all, given an arrow $f\colon X\to Y$ in $\mathsf{Set}$, there is a natural way to turn a subobject of $X$ into a subobject of $Y$: take its image along $f$. Well, this idea only works in a category with a natural notion of "images" of subobjects under arbitrary arrows, e.g. a regular category or an adhesive category. But indeed there are categories where we can make $\text{Sub}$ into a covariant functor.

However, even if we can make $\text{Sub}$ into a covariant functor, we should not typically expect it to be representable. This is because representable functors preserve limits, and the subobject functor rarely preserves limits.

For example, writing $\text{Sub}$ for the covariant subobject functor, if there is an object $\Omega$ such that $\text{Sub}(-)\cong \text{Hom}(\Omega,-)$, then for any product $X\times Y$ in $C$, we have $$\text{Sub}(X\times Y)\cong \text{Hom}(\Omega,X\times Y)\cong \text{Hom}(\Omega,X)\times \text{Hom}(\Omega,Y)\cong \text{Sub}(X)\times \text{Sub}(Y).$$ I hope it's clear that this property is not true in $\mathsf{Set}$ or any other familiar categories (the only natural categories I can think of in which this is true are preorders).

It's much more natural to ask the contravariant subobject functor to be representable. Here, preserving limits means turning colimits in $C$ (limits in $C^{\text{op}}$) into limits in $\mathsf{Set}$. So for example we should have $\text{Sub}(X\sqcup Y)\cong \text{Sub}(X)\times \text{Sub}(Y)$. In $\mathsf{Set}$, this says that picking a subobject of a disjoint union of $X$ and $Y$ is the same as picking a subobject of $X$ and a subobject of $Y$.

Alex Kruckman
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