# Tag Info

6

Although I wouldn't call it an inclusion functor, the answer is yes and in fact the forgetful functor $Set^{C^{op}} \to Set/C_0$ is monadic (as well as comonadic). I think the most illuminating way to see this is to regard a presheaf as a set $F: X \to C_0$ over $C_0$, equipped with a $C$-action which is a map $C_1 \times_{C_0} F \to F$ satisfying suitable ...

3

My paper Non-symmetric $*$-autonomous categories. Theoretical Computer Science, {\bf139} (1995), 115-130, might be thought of as a generalization of Lambek's to Chu categories. It certainly leads to doubly infinite sequences of adjunctions.

4

Another widely used example of infinite adjunction chains arises in linguistics, specifically in connection with pregroup grammars. It has been first observed by Lambek in Some Galois Connections in Elementary Number Theory (J. Number Theory 47 (1994), 371-377). Take monotone maps $f:\mathbb Z\to\mathbb Z$ unbounded in both directions (monotone with respect ...

7

Another nice example is the infinite sequence of adjunctions characterizing stable homotopy theories. One has, in any homotopy theory $K$ (whatever you think that is, as long as there is a notion of (homotopy) limits and colimits) a sequence $1^*\vdash \pi^*\vdash 0^*$ of adjunctions between $K$ and the homotopy theory $K^{[1]}$ of morphisms in $K$, where's $... 16 Broadening the question a bit, you can ask the same question about adjoint 1-morphisms in a 2-category (you're asking about 1-morphisms in the 2-category of categories). Then the 2-dimensional framed bordism 2-category gives a great example which is relatively elementary to understand. Furthermore it "explains" many of the other examples in a sense. A ... 21 Let$C$be a category enriched over finite-dimensional$k$-vector spaces. A Serre functor for$C$is a$k$-linear automorphism$S : C \to C$such that there is a natural equivalence $$\text{Hom}(x, y) \cong \text{Hom}(y, Sx)^{\ast}.$$ Serre functors are unique when they exist. The example that motivates the name occurs when$C = D_b(X)\$ is the bounded ...

Top 50 recent answers are included