Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
added at adjoint functor
more details in the section In terms of universal arrows;
a bit in the section Examples
While $S'\stackrel{S'\eta}\longrightarrow S' TS\stackrel{S'\psi S}\longrightarrow S'T'S\longrightarrow(\epsilon' S}n$Lab is down I must write here some essential notes, which are actually of interest for discussion as well.
Here triple is unfortunately in the sense of monad. Section 3 is called Adjoint triples but is not what we call adjoint triples, but instead the case where the underlying endofunctor of a monad $\mathbf{T} = (T,\mu^T,\eta^T)$ has a right adjoint $G$. Then automatically $G$ is a part of a comonad $\mathbf{G} = (G,\delta^G,\epsilon^G)$ where $\delta^G$ and $\epsilon^G$ are in some sense dual to $\mu^T$ and $\eta^T$. Thus there is a correspondence between monads having right adjoint and comonads having left adjoint, what Rosenberg calls duality. I am not sure that the terminology is optimal. In any case, it is a little more than a consequence of two general facts.
If $T\dashv G$ then $T^k \dashv G^k$ for every natural number $k$.
Given two adjunctions $S\dashv T$ and $S'\dashv T'$ where $S,S': B\to A$, then there is a bijection between the natural transformations $\phi:S'\Rightarrows S$ and natural transformations $\psi:T\Rightarrows T'$ such that
where the horizontal arrows are the natural bijections given by the adjunctions. If $\eta,\eta'$ and $\epsilon,\epsilon'$ are their unit and counit of course the upper arrow is $(SM\stackrel{f}\to N)\mapsto Tf\circ \eta_M$ and the lower arrow $(S'M\stackrel{g}\to N)\mapsto T'g\circ\eta'_M$. Thus the condition renders as
$T'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M$or $T'f\circ T'\phi_M\circ\eta'_M = T'f\circ \psi_{SM}\circ\eta_M$. Given $\phi$, the uniqueness of $B(-,\psi)$ is clear from the above diagram, as the horizontal arrows are invertible. $B(-,\psi)$ determines $\psi$, namely $\psi_N = B(-,\psi)(id_N)$. For the existence of $\psi$ (given $\phi$) satisfying the above equation, one proposes that $\psi$ is the composition $\psi = T'\epsilon \circ T'\phi T \circ \eta'T$, i.e.
$T\stackrel{\eta' T}\longrightarrow T'S' T\stackrel{T'\psi T}\longrightarrow T'ST \longrightarrow{T'\epsilon}\longrightarrow T'$and checks that it works. The inverse is similarly given by the composition
The mechanism strongly reminds of mates, but it is not (classical) mates (in their case one starts with one adjunction). Maybe somebody can elucidate the connection, maybe in some framework it is the same.
This now enables in a special case to dualize $\mu^T$ to $\delta^G$, and similarly unit to the counit. I guess one could do that kind of dualization for more general algebras over operads in the category of endofunctors. By the way, is this extension known ?
No, above is the old form, the newest update has been lost in refresh.
While $S'\stackrel{S' \eta}\longrightarrow S' TS \stackrel{S'\psi S}\longrightarrow S'T'S \stackrel(\epsilon' S}n$Lab is down I must write here some essential notes, which are actually of interest for discussion as well.
Here triple is unfortunately in the sense of monad. Section 3 is called Adjoint triples but is not what we call adjoint triples, but instead the case where the underlying endofunctor of a monad $\mathbf{T} = (T,\mu^T,\eta^T)$ has a right adjoint $G$. Then automatically $G$ is a part of a comonad $\mathbf{G} = (G,\delta^G,\epsilon^G)$ where $\delta^G$ and $\epsilon^G$ are in some sense dual to $\mu^T$ and $\eta^T$. Thus there is a correspondence between monads having right adjoint and comonads having left adjoint, what Rosenberg calls duality. I am not sure that the terminology is optimal. In any case, it is a little more than a consequence of two general facts.
If $T\dashv G$ then $T^k \dashv G^k$ for every natural number $k$.
Given two adjunctions $S\dashv T$ and $S'\dashv T'$ where $S,S': B\to A$, then there is a bijection between the natural transformations $\phi:S'\Rightarrow S$ and natural transformations $\psi:T\Rightarrow T'$ such that
where the horizontal arrows are the natural bijections given by the adjunctions. If $\eta,\eta'$ and $\epsilon,\epsilon'$ are their unit and counit of course the upper arrow is $(SM\stackrel{f}\to N)\mapsto Tf\circ \eta_M$ and the lower arrow $(S'M\stackrel{g}\to N)\mapsto T'g\circ\eta'_M$. Thus the condition renders as
$T'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M$or $T'f\circ T'\phi_M\circ\eta'_M = T'f\circ \psi_{SM}\circ\eta_M$. Given $\phi$, the uniqueness of $B(-,\psi)$ is clear from the above diagram, as the horizontal arrows are invertible. $B(-,\psi)$ determines $\psi$, namely $\psi_N = B(-,\psi)(id_N)$. For the existence of $\psi$ (given $\phi$) satisfying the above equation, one proposes that $\psi$ is the composition $\psi = T'\epsilon \circ T'\phi T \circ \eta'T$, i.e.
$T\stackrel{\eta' T}\longrightarrow T'S' T \stackrel{T'\psi T}\longrightarrow T'ST \stackrel{T'\epsilon}\longrightarrow T'$and checks that it works. The inverse is similarly given by the composition
The mechanism strongly reminds of mates, but it is not (classical) mates (in their case one starts with one adjunction). Maybe somebody can elucidate the connection, maybe in some framework it is the same.
This now enables in a special case to dualize $\mu^T$ to $\delta^G$, and similarly unit to the counit. I guess one could do that kind of dualization for more general algebras over operads in the category of endofunctors. By the way, is this extension known ?
Well now I got it correct in source code. Number 3. Except no formulas in the output!!
More experimenting
$T'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M$Urs’s formula
$\Delta^n \times A^n \simeq \coprod_{\sigma \in A^n} \Delta^n$I am not getting it…
I still cannot see the math in any of the above comments.
Let me try to reinput, that should rerender…I hope.
$T\stackrel{\eta' T}\longrightarrow T'S' T\stackrel{T'\psi T}\longrightarrow T'ST \stackrel{T'\epsilon}\longrightarrow T'$Yes, finally!
Edit: well no. I made some changes to 4 and some formulas now appeared but some did not.
It is strange that the math code from my long comment 4 works when copied and pasted into adjoint monad (new entry announced here).
As far as I can tell from what I can see, this is exactly the notion of mate, in the special case when the functors without adjoints are identities. What do you mean by “in their case one starts with one adjunction”?
Of course, I expected Mike to find the connection.
Hm, now I look into mate and there you indeed start with two adjunctions. I took the notion of mates from Leinster’s book math.CT/0305049 where he starts with one adjunction only, page 150 (180 of the file).
He starts with ONE adjunction $P\dashv Q$, $P: D'\to D$, $Q:D\to D'$ and two functors $T:D\to D$, $T':D'\to D'$. Then there is a correspondence between the transformations $\phi:PT'\Rightarrow TP$ and the transformations $\psi: T'Q\Rightarrow QT$.
The $n$Lab entry mate, as I now see, has two adjunctions and two 1-cells to start with. I did not see that general version. Thanks.
spelled out at adjoint functor in detail the basic proof that a right adjoint is faithful precisely if the counit has epi components.
added at adjoint functor the definition in terms of extensions/liftings, plus when a right/left adjoint is full in terms of the counit/unit
Thanks.
For the sake of those readers who come across this and don’t (yet) know what “$Lan$” stands for and who don’t (yet) know what “absolute” refers to, I have slightly edited to read as follows:
Given $L \colon C \to D$, we have that it has a right adjoint $R\colon D \to C$ precisely if the left Kan extension $Lan_L 1_C$ of the identity along $L$ exists and is absolute, in which case $R \simeq \mathop{Lan}_L 1_C$.
What do you think?
great, thanks. I was in a hurry and didn’t gave a lot of thought to the phrasing. It’s clearer now. I’ve also changed $=$s for $\simeq$s.
Btw, is there a way to link to specific sections within a page? here for example, I’d rather have a link to a (to be added) section of Kan extension detailing when one such is absolute, than to absolute colimit
is there a way to link to specific sections within a page?
Yes, but unfortunately not in a nice wiki-way.
You can link to any URl by writing
[follow this link](http://url.com/see/here.html)
and this may of course include anchors, such as
[follow this link](http://url.com/see/here.html#anchor)
So
[absolute Kan extensiuon](http://ncatlab.org/nlab/show/Kan+extension#AbsoluteKanExtension)
produces the link that you want: absolute Kan extensiuon
(and see the source code of Kan extension for how I produced that anchor to that subsection).
Ok, many thanks Urs
BTW, you can leave off the “http://ncatlab.org” from the URL and it will still work (and even continue to work if the nLab’s domain name had to be changed).
done! added a rel link to the absolute kan extension section
added more historical references
I am trying to bring this page into better expositional shape. First, I have now rewritten and expanded, with full proofs, the section In terms of hom-isomorphisms.
I have also changed the Idea-section. Removed the would-be attempt at a definition, since that is now discussed in the main text. Then I added an absolute minimum of words to put the concept in perspective. Should be expanded. Could be expanded to a long conceptual section, optimally.
Now I have polished and expanded the next section: In terms of representable functors
now I have similarly touched the next section: In terms of universal factorization through a (co)unit
after the lead-in sentence claiming a key role of the concept of adjoint functors, I added this as a footnote:
“In all those areas where category theory is actively used the categorical concept of adjoint functor has come to play a key role.” (first line from An interview with William Lawvere, paraphrasing the first paragraph of Taking categories seriously)
I have polished and expanded the discussion of left adjoints via pointwise limits over comma categories, now this Prop..
replaced the first occurence of an Instikiti array
-hack for adjoint pair notation by a tikzcd
-version
\begin{center}
\begin{tikzcd}
\mathcal{D}
\arrow[r, shift right=6pt, "R"', "\bot"]
&
\mathcal{C}
\arrow[l, shift right=6pt, "L"']
\end{tikzcd}
\end{center}
Suggestions for improvements welcome.
What’s the tikzcd
-analog of what in xymatrix
is
\ar@{<-}[r]
?
\ar[r,<-]
or \ar[from=r]
(I’m not actually sure if these is a difference between these)
Thanks!
I just wanted to add this to the HowTo here, as another escaped-code example, but it seesm that generally the parser gets confused when there is escaped tikz-code, so I am removing it again.
Yes, currently the Tikz code cannot be escaped, either by a <nowiki>
block or in a code block. It is tricky to get this to work due to the interaction between the new and old renderer. I will try to fix it when I get the chance. In the meantime, one can always link to the source.
The issues with escaping should now be fixed.
Fixes typo in diagram: the discussion that follows implies C should be the domain of the functor L and the codomain of the functor R, and not vice versa. There may be other similar typos in the remainder of the article, I haven’t done an exhaustive search (and am not good enough with LaTeX to be confident making extensive edits).
Asad
There is a certain lemma about adjoint functors that makes some of these things more manifest. See “Adjunctions” here https://edeany.com. Are people ok with me adding this lemma here?
Not clear what lemma you’re talking about. You should probably just state your lemma here rather than getting people to copy and paste your address into a separate window and then read a bunch of stuff.
If you write <https://edeany.com>
it will automatically be a link (https://edeany.com).
Is there a typo in diagram (2) (the commutative square)? I don’t see how the vertical arrows are induced by $g:c_2\rightarrow c_1$ and $h:d_1\rightarrow d_2$ - I have read the definition of hom-functor. It works if $Hom(L(c_1), d_1$ is considered instead of $Hom(L(c_2), d_1)$, and same goes for other hom sets.
1 to 58 of 58