Higher category theory tells us that it is a bad idea to identify isomorphic things. Rather, the isomorphism should belong to some additional data. Also, categorification tells us that one should, whenever possible, look at a category directly, not just on its set of isomorphism classes. These are two well-known and accepted principles, right?

However, the definition of a subobject seems to contradict these principles. Why should two monomorphisms be identified with each other when they are isomorphic? What is wrong with the following definition: A subobject of an object $B$ is a monomorphism $A \to B$. With this definition, you can do categorical algebra as usual. It also works nicely in examples. For example, a subring of a ring $B$ is just an injective homomorphism of rings $A \to B$. I think this is more abstract but also more natural than the usual set-theoretic definition; for example with $\mathbb{C}=\mathbb{R}[x]/(x^2+1)$ it is not correct that $\mathbb{R}$ is a subring of $\mathbb{C}$ in the set-theoretic sense, but rather in the sense defined above, via the canonical homomorphism $\mathbb{R} \to \mathbb{C}$. For what purpose should I now look at the class of all monomorphisms $\mathbb{R} \to \mathbb{C}$ isomorphic to that? (This is a bad example since there is only one homomorphism $\mathbb{R} \to \mathbb{C}$ anyway, but I hope that my point is clear. Otherwise consider $\mathbb{Z}[i] \to \mathbb{C}$, $i \mapsto \pm i$.)

  • $\begingroup$ Well still another principle on the next level is that just as in a category the correct notion is isomorphism of objects, not equality, in a 2-category the correct notion is equivalence of objects, not isomorphism. Now subobjects of an object $A$ in a category in your sense themselves form a category, i. e. an object of the 2-category of categories. Subobjects in the old sense form another category. And these two categories are equivalent. So... $\endgroup$ – მამუკა ჯიბლაძე Oct 11 '14 at 21:50
  • $\begingroup$ Yes, I know that. The process basically takes a preorder and makes it a partial order. I think people like partial orders more than preorders - is this the only reason? $\endgroup$ – Martin Brandenburg Oct 11 '14 at 21:52
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    $\begingroup$ I assume the only real reason is that the concept of subobject was invented long before any categorical riff-raff, especially long before higher category theory. At the time it seemed especially perverse to consider any functor representable besides a functor into $Set$. From a modern viewpoint you lose nothing, since the groupoid of monomorphisms is discrete. $\endgroup$ – Anton Fetisov Oct 11 '14 at 22:03
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    $\begingroup$ @AntonFetisov I think your last sentence is wrong in two ways. Firstly, distinguishing discrete groupoids from homotopy discrete groupoids has a long history and many applications. For instance, the coherence theorem of monoidal categories says that certain groupoids are homotopy discrete; but they are certainly not discrete. And the Barratt-Eccles operad is the nerve of an operad in Gpd consisting of homotopy-discrete (indeed, contractible) but not discrete groupoids; if they were replaced by discrete ones then the whole point would be lost. $\endgroup$ – Mike Shulman Jul 15 '17 at 5:06
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    $\begingroup$ Secondly, the whole point of HoTT is that it syntactically enforces homotopy invariance, and thus in particular is unable to distinguish internally between discrete and homotopy-discrete groupoids. $\endgroup$ – Mike Shulman Jul 15 '17 at 5:07

Of course it's not necessary to make this identification, but it's fairly harmless since the groupoid of monomorphisms into an object $X$ is equivalent to the discrete category of subobjects, and it can be a slight technical convenience, especially in relation to smallness conditions. For example, we say that a category is well-powered if for each object the set of its subobjects is "small" (is a set), and this is convenient for example when discussing certain forms of adjoint functor theorems, etc.

In topos theory, assuming that a topos $E$ is well-powered (e.g., a Grothendieck topos), one way of describing a subobject classifier $\Omega$ is that the contravariant subobject functor $Sub: E^{op} \to Set$ is representable as $\hom(-, \Omega)$. This description is conceptually convenient to some people's taste.

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    $\begingroup$ Yes, I know all this, so here are my cons: 1) Many diagrams/categories/limits are actually not small, but only essentially small. So I would define a well-powered category to be one where $\mathrm{Sub}(A)$ is essentially small for every object $A$. (Or even better, define smallness in a better non-set-theoretic way so that this becomes essentially smallness). 2) This should be seen as a functor to categories, or at least, preorders (not partial orders). $\endgroup$ – Martin Brandenburg Oct 11 '14 at 22:03
  • $\begingroup$ (Small error: The category of subobjects is not ess. discrete ...) $\endgroup$ – Martin Brandenburg Oct 11 '14 at 22:05
  • $\begingroup$ (I edited just as you were writing your last comment.) Yes, as I say, you're right -- it's not necessary. But it doesn't seem to be a big deal, as noted already by Anton in a comment above. $\endgroup$ – Todd Trimble Oct 11 '14 at 22:08
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    $\begingroup$ I think the well-poweredness is beside the point: the main thing is that the groupoid core of the the category of subobjects is essentially discrete, and so we are quotienting by unique isomorphisms, which is generally well-behaved. $\endgroup$ – Peter LeFanu Lumsdaine Oct 12 '14 at 0:09
  • $\begingroup$ In fact one might also send both Sub (in the sense of OP) and $hom(-,\Omega)$ to categories (taking $\Omega$ to be an internal poset) and then require these two 2-functors to be equivalent (rather than isomorphic). $\endgroup$ – მამუკა ჯიბლაძე Oct 16 '14 at 9:23

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