Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Enriched categories, topoi, abelian categories, monoidal categories, homological algebra.
7
votes
Accepted
Does the functor $\mathcal{C} \to \mathcal{Z}(\mathcal{C})$ have adjoints?
Short answer: Yes, it can possibly have an adjoint.
Longer answer:
Assume that $\mathcal{C}$ is rigid, and that the coend $L = \int^{X \in \mathcal{C}} X^* \otimes X$ exists.
It is a coalgebra.
Your a …
1
vote
0
answers
126
views
Reference for "adjunction up to twisting by autoequivalences"
Does anyone have any references on the following type of thing, which one might call "adjunction up to autoequivalences"?
We have functors $F \colon C \to D$ and $F' \colon D \to C$, but they are not …