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19 votes

Understanding the tensor-hom adjunction intuitively

arctic tern's answer is very nice and complete. But less formally, the adjunction really says something quite simple: By the universal property of the tensor product, a linear map out of $U \otimes V$ ...
Alex Provost's user avatar
  • 21.1k
16 votes
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The mysterious isomorphism between coinduction and induction

Executive summary: the isomorphism is not particularly mysterious. It arises from the obvious choice of a trace from the group algebra of $G$ to the group algebra of $H$, given by the formula $$\sum_{...
Stephen's user avatar
  • 14.9k
15 votes

Understanding the Beck-Chevalley condition

$\require{AMScd}$ When you do algebraic geometry, you can consider diagrams like $$ \begin{CD} X @>f>> A\\ @VgVV \\ B \end{CD} $$ where $X,A,B$ are spaces and $g,f$ maps thereof. It is ...
fosco's user avatar
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15 votes
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Is there a notion of "schemeification" analogous to that of sheafification of a presheaf?

There is an "affine schemification", a left adjoint to the inclusion of affine schemes into locally ringed spaces. Indeed, given a locally ringed space $X$, the left adjoint just sends $X$ to $\...
Eric Wofsey's user avatar
15 votes
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Why the forgetful functor from $\mathbf{Ab}$ to $\mathbf{Grp}$ does not admit a right adjoint?

Hint: Left adjoints preserve colimits, so it suffices to prove that $U$ does not preserve colimits. For an even bigger hint, hover over the box below.
Clive Newstead's user avatar
15 votes
Accepted

What are some examples of self-adjoint functors? Is this an example?

This is a relatively uncommon scenario, but here are a few more examples. As you suggest, $(-)^\mathrm{op} : \mathbf{Cat} \to \mathbf{Cat}$ is self-adjoint. More generally, this should be true for ...
varkor's user avatar
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13 votes

Let $C,D$ be categories and $F:C\to D$ and $G:D\to C$ be adjoint functors. Then $F$ is fully faithful iff the unit is an isomorphism?

$F$ is fully faithful iff the map $\text{Hom}(x, y) \to \text{Hom}(Fx, Fy)$ is an isomorphism. By adjunction, the map $\text{Hom}(Fx, Fy) \to \text{Hom}(x, GFy)$ is always an isomorphism. Now use the ...
Qiaochu Yuan's user avatar
13 votes

Is the tensor product of two projective sheaves projective?

This fails in general. In fact, it fails for condensed abelian groups, which is a mayor annoyance. Projective condensed abelian groups $M$ are in fact exactly the retracts of $\mathbb Z[S]$ where $S$ ...
Peter Scholze's user avatar
12 votes
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Non-monadic adjunction

A classical example : the forgetful functor from topological spaces to sets. The left adjoint is the "discrete space" functor (sending a set $X$ to the discrete space with underlying space $X$), and ...
Captain Lama's user avatar
  • 26.1k
11 votes

Understanding the tensor-hom adjunction intuitively

There is a de-linearized version of tensor-hom adjunction, called currying in computer science, which states there is a bijection $\hom(U\times V,W)\cong \hom(U,\hom(V,W))$ in the category $\mathsf{...
anon's user avatar
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11 votes
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Quick proof that free objects on sets of different cardinality are not isomorphic?

Counterexample: The Jónsson–Tarski algebra. Let $\mathbf K$ be the class of algebras $\mathfrak A=(A,f,g,h)$ of signature $(2,1,1)$ satisfying the identities $g(f(x,y))=x,\ h(f(x,y))=y,\ f(g(z),h(z))=...
bof's user avatar
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11 votes
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Intuition about coreflective subcategories.

Your intuitive description of reflective subcategories doesn't distinguish left adjoints from right adjoints so you could apply it equally well to either. Anyway, I think the best way to get a handle ...
Qiaochu Yuan's user avatar
10 votes
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Composing functors with natural transformations

Given functors $$\mathcal{A} \xrightarrow{F} \mathcal{B} \overset{G}{\underset{H}{\rightrightarrows}} \mathcal{C} \xrightarrow{K} \mathcal{D}$$ and a natural transformation $\alpha : G \to H$, we can ...
Clive Newstead's user avatar
10 votes
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Hartshorne Exercise II. 1.18

Yes: to prove two morphisms of sheaves are equal, it suffices to prove they induce the same maps on each stalk. Explicitly, let us suppose $F$ and $G$ are sheaves on a space $X$ and $a,b:F\to G$ are ...
Eric Wofsey's user avatar
10 votes

Is “monoidal category enriched over itself” the same as “closed monoidal category”?

The two concepts are distinct. In general it is true that every monoidal closed category is enriched over itself, and this fact is exploited for studying enriched presheaves. The converse does not ...
Giorgio Mossa's user avatar
9 votes
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Adjoints between the category of sets and the category of left G-sets.

The forgetful functor from $G$-sets to sets has adjoints on both sides. The left adjoint sends any set $A$ to the product $G\times A$ with the $G$-action defined by $g(h,a)=(gh,a)$ (for all $g,h\in G$...
Andreas Blass's user avatar
9 votes

Let $C,D$ be categories and $F:C\to D$ and $G:D\to C$ be adjoint functors. Then $F$ is fully faithful iff the unit is an isomorphism?

It's actually possible to prove something a bit more general : $F$ is faithful if and only if every component of $\eta $ is a monomorphism. $F$ is full if and only if every component of $\eta$ is a ...
Arnaud D.'s user avatar
  • 20.9k
9 votes

Show that a functor which preserves colimits has a right adjoint

This is false in general without further hypotheses; see adjoint functor theorem. The reference to CWM, as you say, is only a reference for the uniqueness. The various adjoint functor theorems do ...
Qiaochu Yuan's user avatar
9 votes

Commuting right adjoints implies commuting left adjoints.

There are two results that piece together to give you your answer. Theorem 1. If $\mathcal{C} \overset{F}{\underset{G}{\rightleftarrows}} \mathcal{D} \overset{K}{\underset{H}{\rightleftarrows}} \...
Clive Newstead's user avatar
9 votes
Accepted

What exactly are the `size issues' preventing formation of presheaves being a left adjoint to some forgetful functor?

The problem is that you cannot choose a domain and codomain for such a putative adjunction consistently and simultaneously. The statement that we have is that the category of presheaves on $C$ is the ...
Kevin Carlson's user avatar
9 votes
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Right adjoint is fully faithful iff the counit is an isomorphism (without Yoneda)

This result uses the naturalness of the map $$ \Phi \colon \mathrm{Hom}(A,GB) \xrightarrow\sim \mathrm{Hom}(FA,B). $$ In particular, if $f\colon A\to GB$ and $g\colon B\to B’$, then $\Phi(G(g)f)=g\Phi(...
Andrew Hubery's user avatar
9 votes
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Explicit description of a left adjoint to the forgetful functor from Unif to Top

It does not follow from the existence of a left adjoint $F\colon \mathsf{Top}\to \mathsf{Unif}$ that every topological space has a canonical uniformity compatible with its topology. There's no reason ...
Alex Kruckman's user avatar
9 votes
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Bifunctoriality stronger than functoriality in each variable?

Yes, bifunctoriality is a stronger condition than functoriality in each variable, and yes, the commutative diagram you write down need not commute in the second case. I don't actually know an example ...
Qiaochu Yuan's user avatar
9 votes

Left adjoint to the forgetful functor from Hilbert spaces to topological spaces

The left adjoint does not exist. Otherwise, also the forgetful functor $\mathbf{Hilb} \to \mathbf{Top} \to \mathbf{Set}$ to the category of sets had a left adjoint (since left adjoints "compose&...
Martin Brandenburg's user avatar
8 votes
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What can be said about adjunctions between groups (regarded as one-object categories)?

Since groups as one-object categories are a special case of groupoids, that is, categories in which every morphism is an isomorphism, for any adjunction $F\dashv H\colon\mathcal G_1\to\mathcal G_2$ ...
Vladimir Sotirov's user avatar
8 votes
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Uniqueness of Left Adjoint

The left adjoint is unique up to natural isomorphism. To see this, let $G : \mathcal{D} \to \mathcal{C}$ be a functor and suppose $F,F' : \mathcal{C} \to \mathcal{D}$ with are two left adjoints for $G$...
Clive Newstead's user avatar
8 votes
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Is the free group functor an isomorphism?

OK, so you can do that to convert a set to a group. Now take that group and apply the forgetful functor: what set do you get? In particular, if $S = \{a\}$, then $G$ looks like the group $\Bbb Z$. ...
John Hughes's user avatar
8 votes
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Does the forgetful functor $U: \mathbf{Gph} \to \mathbf{Set}$ have a left adjoint?

Whenever I say graph in this answer, I mean simple graph. We need to be a bit more precise here, and specify what the arrows are in $\mathbf{Gph}$. A natural choice would be maps $f: G \to G'$ ...
Mark Kamsma's user avatar

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