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Stream: learning: reading & references

Topic: Inverting to adjoint isomorphisms

Fernando Yamauti (Jan 14 2025 at 19:45):

Hi. Does anyone know if anyone has already studied free inversions of arrows into adjoint isomorphisms?

More precisely, it's known that the walking isomorphism of $(\infty, n)$-categories is not contractible, but it's variant with more additional data, the walking adjoint isomorphism, is supposed to always be contractible (https://arxiv.org/abs/2303.00567). On the other side, it seems that it's always the case one can individually upgrade isomorphisms to adjoint ones.

Now, one can consider a variant of localisation for some class of $1$-cells $W$ in a $(\infty, n)$-category $C$, $C [W^{-adj}]$, such that every arrow in $W$ becomes, instead, an adjoint isomorphism.

Has anyone ever studied before this variant of localisation? I'm particularly interested in conditions guaranteeing $C [W^{-adj}] \cong C [W^{-1}]$ (maybe that's almost never the case, though).

Mike Shulman (Jan 14 2025 at 22:18):

The standard meaning of $C[W^{-1}]$ will always be equivalent to your $C[W^{-\mathrm{adj}}]$, because it refers to forcing 1-cells to have the property of being an isomorphism rather than the structure of a non-adjoint isomorphism.

Fernando Yamauti (Jan 14 2025 at 22:42):

Mike Shulman said:

The standard meaning of $C[W^{-1}]$ will always be equivalent to your $C[W^{-\mathrm{adj}}]$, because it refers to forcing 1-cells to have the property of being an isomorphism rather than the structure of a non-adjoint isomorphism.

I'm confused. I feel like I'm overseeing something extremely trivial. Let's restrict everything to $(2, 1)$-cats. Shouldn't $\Delta^1[\Delta^{-1}]$ and $\Delta^1[\Delta^{-adj}]$ be respectively the walking isomorphism and the walking adjoint isomorphism? If so, the latter is always supposed to be equivalent to the terminal category, while the former cannot satisfy so according to Prop. 1.3.2 of the aforementioned paper...

Mike Shulman (Jan 15 2025 at 15:00):

No, as I said, $\Delta^1[(\Delta^1)^{-1}]$ means $\Delta^1$ with the morphism freely forced to have the property of being an isomorphism, which is equivalent to having the structure of an adjoint isomorphism.

Fernando Yamauti (Jan 15 2025 at 18:01):

Mike Shulman said:

No, as I said, $\Delta^1[(\Delta^1)^{-1}]$ means $\Delta^1$ with the morphism freely forced to have the property of being an isomorphism, which is equivalent to having the structure of an adjoint isomorphism.

Hmm... Let me see if I've got it. So the point is that every pair of $1$-cells $f$ and $g$ together with $2$-cells $fg \cong 1$ and $gf \cong 1$ can be modified to another one by changing the $2$-cells so that the triangle identities are true. And that doesn't contradict the non-contractibility of the walking iso because liftings from an isomorphism datum to an adjoint isomorphism datum are not unique (if they were unique, the walking iso would be equivalent to the walking adjoint iso). Did I interpret your reply properly?

But, then, it seems that the usual localisation formula, where one takes pushouts along coproducts of the inclusions into walking isos $\Delta^{1} \rightarrow \Delta^{1}_{\text{iso}}$ can be instead computed using $\Delta^{1} \rightarrow \Delta^{1}_{\text{adj-iso}} \cong 1$... I'm skeptic that's always the case. Maybe I'm overseeing something again?

Mike Shulman (Jan 15 2025 at 18:12):

Fernando Yamauti said:

But, then, it seems that the usual localisation formula, where one takes pushouts along coproducts of the inclusions into walking isos $\Delta^{1} \rightarrow \Delta^{1}_{\text{iso}}$

I don't know where you saw that, but that is not the correct localization formula when working with $n$-categories for $n>1$; you do have to use the walking adjoint isomorphism. My point was that this is what is generally meant by the notation $C[W^{-1}]$.

Mike Shulman (Jan 15 2025 at 18:14):

In other words, the version where you force things to become non-adjoint isomorphisms is the one that's a "variant" of localization, which I doubt anyone has studied very much.

Fernando Yamauti (Jan 15 2025 at 18:47):

Mike Shulman said:

I don't know where you saw that, but that is not the correct localization formula when working with $n$-categories for $n>1$; you do have to use the walking adjoint isomorphism. My point was that this is what is generally meant by the notation $C[W^{-1}]$.

Wait. I recall seeing written somewhere (I'm still trying to find it) that the simplicial localisation of a $1$-cat could be computed using homotopy pushouts (in the Joyal model structure) along (the nerve of) inclusions of $\Delta^1$'s into walking $1$-isomorphisms (everything is a $1$-cat here). That can't be correct then?

Mike Shulman (Jan 15 2025 at 18:51):

Ah! The nerve of the "walking isomorphism" 1-category is the walking adjoint $(\infty,1)$-isomorphism! This is kind of magic, but it's true.

Fernando Yamauti (Jan 15 2025 at 19:28):

Mike Shulman said:

Ah! The nerve of the "walking isomorphism" 1-category is the walking adjoint $(\infty,1)$-isomorphism! This is kind of magic, but it's true.

So that's the source of all my confusion until now! Btw, is that a general pattern? That is: for whatever geometric nerve from $(m, m)$-cats and any $n$, should it always be the case that applying the nerve and truncating to the homotopy $(n, n)$-cat sends the "walking iso" $m$-cat to the "walking adjoint iso" $n$-cat? That seems to work for the Duskin nerve...

Mike Shulman (Jan 15 2025 at 19:43):

I would say that this operation should send the walking adjoint isomorphism $m$-category to the walking adjoint isomorphism $n$-category. It's just that for 1-categories, there is no difference between isomorphisms and adjoint isomorphisms.

Fernando Yamauti (Jan 15 2025 at 20:10):

Mike Shulman said:

I would say that this operation should send the walking adjoint isomorphism $m$-category to the walking adjoint isomorphism $n$-category. It's just that for 1-categories, there is no difference between isomorphisms and adjoint isomorphisms.

Ah! Ok. Thanks. Duh! Of course, if the triangle identities are not true, then there's no $3$-cell in the Duskin nerve filling the boundary of the tetrahedron with two degenerated faces and one face for the (co)unity. So nothing weird happens as expected and all the usual nerves are good models for inclusions into higher dimensions (which should always be conservative, so no non-contractible object can become contractible).

Fernando Yamauti (Jan 16 2025 at 18:39):

Let me just add something that summarise why $\Delta^1 [(\Delta^{1})^{-1}] \cong \Delta^1_{\text{adj-iso}}$. @Mike Shulman , let me know if I'm not spilling any bullshit.

The point is: if $a \colon \Delta^1 \rightarrow C$ defines an arrow that happens to be invertible, then the space of choices of liftings of $a$ along the canonical inclusion $\Delta^1 \rightarrow \Delta^1_{\text{iso}}$ is not contractible, while there is exactly one unique lifting of $a$ (up to contractible space of choices) $\Delta^1_{\text{adj-iso}} \rightarrow C$ along the inclusion $\Delta^1 \rightarrow \Delta^1_{\text{adj-iso}}$. And all that is simply equivalent to the contractibility of $\Delta^1_{\text{adj-iso}}$!

That clarifies a fundamental misunderstanding I had regarding localisations of higher stuff.

Mike Shulman (Jan 16 2025 at 19:00):

Yes, that's right.