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Stream: deprecated: logic

Topic: linear version of Yoneda?

Lucius Gregory Meredith (Jun 20 2022 at 20:31):

Naively, i think of Yoneda as saying we can get a view of an object X in some category C by collecting into a set all the maps in C with target X. We could modify this procedure by 1) demanding that the maps we collect pass a filter; and 2) collecting the results in something other than a set.

i am particularly interested in the case where we demand that the maps are linear in some suitable sense and that the resulting maps are collected in a “linear structure”. Is there already a version of Yoneda that generalizes in this way? I know that there are enriched versions of Yoneda, which covers the second requirement, but what about the first?

Simon Burton (Jun 20 2022 at 22:03):

(1) this would be Yoneda lemma for a subcategory, right?

Matteo Capucci (he/him) (Jun 21 2022 at 21:44):

Lucius Gregory Meredith said:

the maps are linear in some suitable sense

I'm afraid you might need to be more specific! If you mean 'linear' as in 'linear algebra', then you're arguing Yoneda should be true if your category is a category of linear maps (like a category of modules, vector spaces or abelian groups), and that's surely the case.

Alexander Gietelink Oldenziel (Nov 08 2022 at 15:02):

The 2-category of categories is an equipment. Another important equipments is the cateory of k- rings, ring morphisms and bi-modules, where $k$ is some ring. The presheaf category $Psh(C)$ in Cat can also be considered as bimodules/profunctors $H: C \to 1$. If we extend this analogy, the 'presheafs' on an object $R$ should be R-k bi-modules. What should be the Yoneda embedding? I'd conjecture it should correspond to embedding the ring $R$ as the canonical R-k bimodule... but I haven't checked this.

John Baez (Nov 08 2022 at 16:00):

I think the analogy to ordinary category works quite smoothly if instead of just $k$-rings, usually called algebras over the commutative ring $k$, we consider $k$-ringoids, which are usually called $k$-algebroids. These are just enriched categories of a certain sort, namely categories enriched over $\mathsf{Mod}_k$: the symmetric monoidal category of $k$-modules with its usual tensor product.

John Baez (Nov 08 2022 at 16:01):

In the case $k = \mathbb{Z}$ they are called ringoids, or more often Ab-enriched categories.

John Baez (Nov 08 2022 at 16:02):

A $k$-ring in your sense, i.e. an algebra over $k$, is the same as $k$-algebroid with one object.

John Baez (Nov 08 2022 at 16:05):

I believe we can then see your equipment is a 'sub-equipment' of the double category of

• categories enriched over $\mathsf{Mod}_k$
• enriched functors
• enriched profunctors

John Baez (Nov 08 2022 at 16:05):

For example a bimodule between $k$-algebras is an enriched profunctor between the corresponding one-object $k$-algebroids.

John Baez (Nov 08 2022 at 16:07):

My reason for introducing all this nonsense is that we should then be able to use some sort of more or less familiar Yoneda embedding for enriched categories to think about your proposed Yoneda embedding.

John Baez (Nov 08 2022 at 16:08):

And I think the way it works is that any $k$-algebroid $A$ embeds into the category of enriched presheaves on $A$.

John Baez (Nov 08 2022 at 16:08):

These enriched presheaves are enriched functors $A^{\text{op}} \to \mathsf{Mod}_k$.

John Baez (Nov 08 2022 at 16:10):

And when $A$ is a one-object $k$-algebroid - that is, an algebra over $k$ - these enriched functors are just right $A$-modules.

John Baez (Nov 08 2022 at 16:11):

And I think the way the enriched Yoneda embedding works in this case is that the algebra $A$ gets embedded in the algebra of right $A$-module maps $A \to A$.

John Baez (Nov 08 2022 at 16:12):

Namely, $a \in A$ gets sent to left multiplication $L_a : A \to A$, which is a right $A$-module map.