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Stream: theory: category theory

Topic: Fullness on isomorphisms

Amar Hadzihasanovic (Jan 06 2021 at 14:08):

I have a faithful forgetful functor $U: C \to D$ and I'm trying to capture the idea that “a $D$-object admits, essentially, at most one structure of $C$-object”, so an isomorphism-robust version of injective-on-objects. The correct notion, in my case, seems to be that, in addition to being faithful, $U$ reflects and is full on isomorphisms, that is, any isomorphism $f': Uc \to Ud$ is the image of a (necessarily unique) isomorphism $f: c \to d$.

I have the impression that this kind of forgetful functors is quite common: examples (if I'm not wrong) are the forgetful functor $\mathbf{Mnd} \to \mathbf{SemiGp}$ from monoids to semigroups, or the forgetful functor $\mathbf{CartMonCat} \to \mathbf{MonCat}$ from (cartesian monoidal categories, strong symmetric monoidal functors) to (monoidal categories, strong monoidal functors)...
So I would imagine that the notion deserves a name, but I don't think I've seen it before. Is this familiar to anyone?

Amar Hadzihasanovic (Jan 06 2021 at 14:15):

(I think my second example is probably just full, but the first should be fine, and I can think of a few more...)

Tobias Fritz (Jan 06 2021 at 14:19):

Hi, this sounds morally related to this MO discussion, although the precise condition mentioned there is a little different.

Nathanael Arkor (Jan 06 2021 at 14:21):

This is not quite the same, but it is perhaps suitable for your purposes. A monad on a complete category is property-like if the forgetful functor $U : T\text{-}\mathrm{Alg} \to \mathscr C$ is injective-on-objects, or equivalently if it is pseudomonic (i.e. faithful and full on isomorphisms). (This is Theorem 7.1 of Kelly–Lack's On property-like structures.)

Nathanael Arkor (Jan 06 2021 at 14:22):

This just lacks your isomorphism reflection condition.

Amar Hadzihasanovic (Jan 06 2021 at 14:30):

Ah, but pseudomonicity implies isomorphism reflection. If $Uf$ is an isomorphism, then it lifts to some isomorphism $g$, and $Uf = Ug$ implies $f = g$ by faithfulness, so $f$ was an iso to begin with.

Amar Hadzihasanovic (Jan 06 2021 at 14:30):

So that's exactly what I was looking for. Thank you Nathanael!

Nathanael Arkor (Jan 06 2021 at 14:31):

Ah, of course! Perfect :)

Amar Hadzihasanovic (Jan 06 2021 at 14:33):

(Thanks Tobias too -- that was relevant, but “creates isomorphisms” does seem to be subtly different, that is, it just requires that if there is an iso $f: Uc \to Ud$, there is some iso $g: c \to d$, but not necessarily $Ug = f$...)

Tobias Fritz (Jan 06 2021 at 14:58):

Right, that's why I only said that it was "morally" related :smile: for a category theorist clearly your version is the more natural one.