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Let us say that a category $\mathcal C$ has enough of some class $\mathcal U$ of object if every object in $\mathcal C$ is a colimit of objects of the class $\mathcal U$. The pointset topology analogue of enoughness is density: a subset $U$ of a topological space $C$ is dense if every point in $C$ is a limit of points in the $U$.

Two dense sets may have empty intersection ($\mathbb Q \subset \mathbb R$ and $\mathbb Q + \pi \subset \mathbb R$, for example), but the intersection of dense open sets is always dense and open.

What is the analogue of openness for categories? For example, in $\mathrm{Vec}$, there are enough 5-dimensional objects, and also enough 6-dimensional objects, but there are not enough objects which are simultaneously 5- and 6-dimensional, and so 5-dimensionality is not "open". But I'm pretty sure that in any abelian category, if there are enough projective objects and also enough compact objects, then there are enough objects which are simultaneously compact and projective (right?), and so I expect compactness and projectivity are "open" conditions.

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  • $\begingroup$ It's worth reminding that categories have two sometimes-competing "topologies", because they have both limits and colimits. I've somewhat arbitrarily chosen the colimit topology, but note that when an abelian category has, say, "enough injectives", then that's a limit-topology statement, not a colimit-topology statement. $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 8, 2019 at 17:00
  • $\begingroup$ Perhaps another example is accessible subcategories of accessible categories, with the compact objects as the dense subset. Accessible subcategories agree about which objects are compact, they are closed under intersection, and the preimage of an accessible subcategory under an accessible functor is also an accessible subcategory. This involves restricting to filtered colimits, though. $\endgroup$
    – John Dougherty
    Commented Sep 8, 2019 at 18:00
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    $\begingroup$ There are several different kinds of "enoughness" / "density" and they don't always agree, e.g. you might want iterated colimits rather than just colimits or else $\mathbb{Z}$ isn't dense in abelian groups. See e.g. qchu.wordpress.com/2015/05/17/generators for a discussion. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 8, 2019 at 19:50
  • $\begingroup$ @QiaochuYuan Good point. But permit me to be open-minded? I'm often annoyingly imprecise :) $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 9, 2019 at 16:11

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Unsurprisingly, I think there will be more than one way to answer this question.

Somewhat more precisely, I think the answer will hinge upon which of two (or more!) ways we choose to view a category as being like a space.

Analogy 1: A category $C$ is like a directed space $X$ where a morphism of $C$ is a directed path in $X$.

There's an amusingly literal way to draw out this analogy. Recall that, following Lawvere, a category enriched in the monoidal poset $([0,\infty], \geq , +)$ is a sort of generalized metric space (where the distance may be asymmetric and distinct points may be distance 0 from each other, but the triangle inequality holds). So more generally, one may think of a $V$-enriched category as being a sort of space with a $V$-object's worth of paths, and generalize geometric concepts from the metric space case.

Open sets will correspond to presheaves (i.e. 1-Lipschitz functions), with the open set being the preimage of the complement of 0 under the functor / function. So this turns out to just consist of sieves on the category $C$, which is perhaps a bit disappointing.

Analogy 2: A category $C$ is like a space $X$ where an object of $C$ is a sheaf over $X$.

I imagine that if you're working in a linear world, you really want to think of the objects of your categories as being some kind of quasicoherent sheaf, so this may be a better way to think about things.

In this case, an etale map to $X$ should correspond to some kind of sheaf $F$ on $C$. The total space of the sheaf $F$ is the Grothendieck construction / category of elements of $F$. An open set should correspond to an etale map such that the forgetful functor from the category of elements is fully faithful, i.e. to a subterminal object, which is perhaps again a bit disappointing. Maybe this isn't so surprising though, as very often one doesn't really want to work with the Zariski topology anyway.

Probably what one should really do is customize these analogies (and others) to specific situations. There will also be other ways of drawing out these (and other) analogies.

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