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Forgive me if this question turns out to be too elementary-then feel free to move it to stack exchange. I believe that this should be very basic fact from topos theory nevertheless being not familiar with topos theory let me ask it. One of the highlights of topos theory is the possibility to consider an analogue of the power set in the form of $Sub(A)=\{all \ subobjects \ of \ A \}$ (which are defined as equiavlence classes of monic arrows). In any topos there is a distinguished object $\Omega$ called the subobjects classifier. Moreover in any topos there is an analogue of the operation of taking all functions from $A \to B$-this is called exponentation. Given those facts we have a one to one correspondence $Sub(A) \cong \Omega^{A}$. Am I right in saying that:

The left hand side of this correspondence is a priori not an object of our topos $\mathcal{T}$ but the right hand side is: therefore this natural correspondence is the way of making $Sub(A)$ again an object in our category $\mathcal{T}$

Please corect me if I'm wrong

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2 Answers 2

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The general notion you're looking for is a representable functor. For example:

$${\mathcal E}(X\times{-},Y) \sim {\mathcal E}({-},Y^X)$$

$${\textsf{Sub}}({-}) \sim {\mathcal{E}}({-},\Omega)$$

The thing on the left is a general contravariant functor from the category to $\mathbf{Set}$.

The thing on the right is of the form ${\mathcal E}({-},R)$, where $R$ is the "representing object" for the functor.

The definition of a préfaisceau représentable is given in SGA4 Exposé I, remark 1.4.2. Presumably the idea is then used, maybe extensively, in SGA4 and beyond, but I will leave others to search for it.

This is also one of the many equivalent ways of expressing an adjunction.

Of course, whether or not the (topos or other) category has a representing object for a particular functor is very much up for debate in the topic in question.

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  • $\begingroup$ Thank you-as I have understood, in your answer if we plug in some $A$ to both sides we get just a set: but I would like $Sub(A)$ to be an object in my topos $\endgroup$
    – truebaran
    Commented Sep 15, 2021 at 16:48
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    $\begingroup$ Paul gave you all the ingridients: $Sub(A) \sim \mathcal{E}(A, \Omega) \sim \mathcal{E}(1 \times A, \Omega) \sim \mathcal{E}(1, \Omega^A)$, so the object you are looking for is $\Omega^A$ (because the global points of $\Omega^A$ are in natural bijective correspondence with the subobjects of $A$). $\endgroup$ Commented Sep 15, 2021 at 16:53
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To 'turn an object $A$ of a category $\mathcal{C}$ into a set' you can consider the set of objects mapping into/out of that object from/to some other object; if you consider maps out of a terminal object ${\bf 1}\to A$ these are called global elements of the object.

If $\mathcal{C}$ is a well-pointed topos these global elements are sufficient to delineate between parallel arrows with domain $X$ and thusly contain 'all relevant information' about $X$ in the topos. As Paul pointed out in his answer and Andrej elaborated on in his comment, if you consider $\Omega^A$ as a set in this way by considering its set of global elements $\{f:{\bf 1}\to \Omega^A|f\in\mathcal{C}\}$, this set is isomorphic to $Sub(A)$ as a lattice.

You may also want to read about the subobject fibration, which deals with the notion of categorical subobjects in a more satisfying way imo; you can turn the subobjects of an object into a category in an obvious way, and it turns out these categories are also the fibers of the fibration obtained by turning the codomain fibration into an indexed category via the Grothendieck construction, postcomposing with the skeleton endo $2$-functor on $\mathfrak{Cat}$, then turning the resulting indexed category back into a fibration via Grothendieck in the other direction.

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