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I'm looking for a reference for the following extension property: A functor $F: C \to D $ such that for any arrow $f: x \to y$ in $C$ there is an extension $f = \hat{f} \circ i$ with $i:x \hookrightarrow z$ monic and $F\hat{f}$ an isomorphism in $D$.

My motivation is that if $D$ is an ordinal category then this is a really simple (and restrictive) definition of dimension that fits other needs of my research.

EDIT: Expanding on the dimension part: the point to use this as a definition of dimension has several advantages:

  1. The dimension of pullbacks/pushouts acts as expected.

  2. There are no morphisms from a higher dimension object to a lower dimension one.

This means that the dimension property is, in some sense, external to the underlying categorial structure. If we view this dimension structure as geometrical - it helps separate geometric constructions from the underlying building blocks of the construction - the category itself (this also creates difficulties modeling this dimension functor in concrete cases, but that is a technical issue in my point of view).

The last point can be extended further: The existence of such functor can be used to restrict the behavior of the morphisms related to this "external structure" (for ordinal $D$ this is point 2). The extension property is there to connect the "external structure" to the underlying category, but in a very controlled way. Now, dimension is not necessarily the only thing we want to control.

For example we can take $D=\alpha\times E$ with the first coordinate meaning dimension, and the second a discrete category - this is for any invariant under existence of morphisms (say Euler characteristic). The existence of a functor, both restricts what morphisms are allowed - only increasing the dimension and preserving some strong invariant. Every morphism can be studied as an embedding into the $D$-isomorphism class of it's codomain.

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  • $\begingroup$ You don't usually ask for a solution to a lifting problem to belong to a certain class of maps (those $f$'s inverted by $F$ are called $F$-local maps); instead, you ask it to the "sides" of the square. This to say that I've never seen this before (but I'll gladly learn something new!) $\endgroup$
    – fosco
    Aug 1, 2018 at 18:36
  • $\begingroup$ Ah, another clarification in case I have something more to say: is it enough for you to consider $D$ an ordinal? This is a great simplification! $\endgroup$
    – fosco
    Aug 1, 2018 at 18:37
  • $\begingroup$ @FoscoLoregian Well, right now ordinals is all I need. But, hypothetically you could replace this ordinal with a different category encoding both dimension and Euler characteristic of the space, or any other interesting invariants you want. So this might be of interest to explore different categories as well $\endgroup$ Aug 1, 2018 at 21:17
  • $\begingroup$ A few things: 1. What does it mean for pullbacks to behave "as expected"? 2. What is the significance of the extension property? Point (2) follows simply from the existence of a functor $F: C \to D$ and point (1) follows simply from $F$ preserving pushouts (though I'm confused about pullbacks). 3. What's an example? For instance, consider the category of sets and all functions $X \to Y$ where $|X| \leq |Y|$. Then the $Ord$-valued functor $F(X) = max(|X|^+,\aleph_0)$ preserves pushouts (but not pullbacks) and has the extension property. Is this the sort of thing you have in mind? $\endgroup$
    – Tim Campion
    Aug 4, 2018 at 21:02
  • $\begingroup$ @TimCampion To answer points 1 and 2: If we look at pullbacks in an over category, then they correspond to intersection. With this intuition, if we look at every object as a subobject of object of higher dimension (the extension property is this exactly) - the dimension of "intersection" of subobjects is $\leq$ the dimension of the objects. The extension property is there to make this intuition rigorous $\endgroup$ Aug 5, 2018 at 18:36

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