The adjoint-functors tag has no wiki summary.

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### “adjoint” =?= “inverse of composite endofunctor is uniform bi-composition”

Understanding adjoints has always been (and continues to be) a bit of a struggle for me.
Today I stumbled upon a property of adjoint functors which seemed extremely intuitive to me. I was wondering ...

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### Left adjoint Grp->Set??

Was wondering if someone could help. I am just about getting to grips with cateogry theory and adjoints. I have a query though, reading through some material.
Does the forgetful functor from the ...

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### Iterated adjoint functors

Let $F_0 : C \to D$ be a functor. If it exists, let $G_0 : D \to C$ be its left adjoint. If it exists, let $F_1 : C \to D$ be its left adjoint. And so forth. In situations where the infinite ...

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### Is a functor which has a left adjoint which is also its right adjoint an equivalence ?

I am looking for a counter-example of two functors F : C -> D and G : D->C such that
1) F is left adjoint to G
2) F is right adjoint to G
3) F is not an equivalence (ie F is not a quasi-inverse of ...

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### How to construct pair of adjoint functors from category A to category A_D(category of diagrams)

I wonder whether following statements holds
If A is an abelian category(or quasi abelian category) having enough projectives, then category of pointed diagram(which means diagram has final object,or ...

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### When and why do universal objects have extra properties?

I'm interested in situations where universal objects come with more structure than their definitions suggest. A classic case of this is where the free abelian group on one element has a ring ...

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### What is an intuitive view of adjoints? (version 1: category theory)

In trying to think of an intuitive answer to a question on adjoints, I realised that I didn't have a nice conceptual understanding of what an adjoint pair actually is.
I know the definition (several ...

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### Can different modules have the same symmetric algebra? (answered: no)

Algebraic geometry allows one to think of an $A$-module $M$ geometrically as a module of functions on the $A$-scheme $\mathrm{Spec}(\mathrm{Sym}(M))$.
I'm wondering if anything is lost in just ...

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### What is the “right” definition of the free abelian group on a set?

One way to define the free abelian group on a set $S$ is as $F(S) := \mbox{Hom}_{\text{Set}}(S, \mathbb{Z})$ with the group operation coming by composition with the map $\mathbb{Z} \times \mathbb{Z} ...

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### How do I check if a functor has a (left/right) adjoint?

Because adjoint functors are just cool, and knowing that a pair of functors is an adjoint pair gives you a bunch of information from generalized abstract nonsense, I often find myself saying, "Hey, ...

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### What's an example of an “adjunction up to adjunction”?

(There are several dialects of 2-categorical language. Mine is the one where everything is weak by default. You might need to insert the word "weak" or "pseudo" (or even "strong", if your default is ...

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### When does the sheaf direct image functor f_* have a right adjoint?

Say f: X → Y is a morphism of schemes. The sheaf direct image functor f★ always has a left adjoint, namely the sheaf inverse image functor f★ (with tensoring).
Under what ...

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### Is there a free digraph associated to a graph?

A little bit of background: A graph G is, of course, a set of vertices V(G) and a multiset of edges, which are unordered pairs of (not necessarily distinct) vertices. We say that two vertices v_1, v_2 ...

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### Can adjoint linear transformations be naturally realized as adjoint functors?

Last week Yan Zhang asked me the following: is there a way to realize vector spaces as categories so that adjoint functors between pairs of vector spaces become adjoint linear operators in the usual ...

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### Does a scheme have a “separification”?

Background:
(1) If C and D are categories and there is a forgetful functor U:C→D, then a C-ification functor F:D→C is an adjoint to U. For example, the (left) adjoint to the forgetful ...

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### Is every functor a composition of adjoint functors?

My understanding of Ben's answer to this question is that even though associated graded is not an adjoint functor, it's not too bad because it is a composition of a right adjoint and a left adjoint.
...