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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeApr 30th 2010

    added at adjoint functor

    • CommentRowNumber2.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    • (edited Apr 24th 2011)

    While n S'\stackrel{S'\eta}\longrightarrow S' TS\stackrel{S'\psi S}\longrightarrow S'T'S\longrightarrow(\epsilon' S}nLab is down I must write here some essential notes, which are actually of interest for discussion as well.

    • Samuel Eilenberg, John C. Moore, Adjoint functors and triples, Illinois J. Math. 9, Issue 3 (1965), 381-398.

    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 T=(T,μ T,η T)\mathbf{T} = (T,\mu^T,\eta^T) has a right adjoint GG. Then automatically GG is a part of a comonad G=(G,δ G,ε G)\mathbf{G} = (G,\delta^G,\epsilon^G) where δ G\delta^G and ε G\epsilon^G are in some sense dual to μ T\mu^T and η T\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.

    1. If TGT\dashv G then T kG kT^k \dashv G^k for every natural number kk.

    2. Given two adjunctions STS\dashv T and STS'\dashv T' where S,S:BAS,S': B\to A, then there is a bijection between the natural transformations ϕ:SRightarrowsS\phi:S'\Rightarrows S and natural transformations ψ:TRightarrowsT\psi:T\Rightarrows T' such that

    A(S,) B(,T) A(ϕ,) B(,ψ) A(S,) B(,T)\array{ A (S,-) &\to& B(-,T)\\ A(\phi,-)\downarrow &&\downarrow B(-,\psi)\\ A(S',-)&\to & B(-,T') }

    where the horizontal arrows are the natural bijections given by the adjunctions. If η,η\eta,\eta' and ε,ε\epsilon,\epsilon' are their unit and counit of course the upper arrow is (SMfN)Tfη M(SM\stackrel{f}\to N)\mapsto Tf\circ \eta_M and the lower arrow (SMgN)Tgη M(S'M\stackrel{g}\to N)\mapsto T'g\circ\eta'_M. Thus the condition renders as

    T(fϕ M)η M=ψ NTfη MT'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M

    or TfTϕ Mη M=Tfψ SMη MT'f\circ T'\phi_M\circ\eta'_M = T'f\circ \psi_{SM}\circ\eta_M. Given ϕ\phi, the uniqueness of B(,ψ)B(-,\psi) is clear from the above diagram, as the horizontal arrows are invertible. B(,ψ)B(-,\psi) determines ψ\psi, namely ψ N=B(,ψ)(id N)\psi_N = B(-,\psi)(id_N). For the existence of ψ\psi (given ϕ\phi) satisfying the above equation, one proposes that ψ\psi is the composition ψ=TεTϕTηT\psi = T'\epsilon \circ T'\phi T \circ \eta'T, i.e.

    TηTTSTTψTTSTTεT 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 μ T\mu^T to δ G\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 ?

    • CommentRowNumber3.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    • (edited Apr 24th 2011)

    No, above is the old form, the newest update has been lost in refresh.

    • CommentRowNumber4.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    • (edited Apr 25th 2011)

    While n S'\stackrel{S' \eta}\longrightarrow S' TS \stackrel{S'\psi S}\longrightarrow S'T'S \stackrel(\epsilon' S}nLab is down I must write here some essential notes, which are actually of interest for discussion as well.

    • Samuel Eilenberg, John C. Moore, Adjoint functors and triples, Illinois J. Math. 9, Issue 3 (1965), 381-398.

    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 T=(T,μ T,η T)\mathbf{T} = (T,\mu^T,\eta^T) has a right adjoint GG. Then automatically GG is a part of a comonad G=(G,δ G,ε G)\mathbf{G} = (G,\delta^G,\epsilon^G) where δ G\delta^G and ε G\epsilon^G are in some sense dual to μ T\mu^T and η T\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.

    1. If TGT\dashv G then T kG kT^k \dashv G^k for every natural number kk.

    2. Given two adjunctions STS\dashv T and STS'\dashv T' where S,S:BAS,S': B\to A, then there is a bijection between the natural transformations ϕ:SS\phi:S'\Rightarrow S and natural transformations ψ:TT\psi:T\Rightarrow T' such that

    A(S,) B(,T) A(ϕ,) B(,ψ) A(S,) B(,T) \array{ A (S,-) &\to& B(-,T)\\ A(\phi,-)\downarrow &&\downarrow B(-,\psi)\\ A(S',-)&\to & B(-,T') }

    where the horizontal arrows are the natural bijections given by the adjunctions. If η,η\eta,\eta' and ε,ε\epsilon,\epsilon' are their unit and counit of course the upper arrow is (SMfN)Tfη M(SM\stackrel{f}\to N)\mapsto Tf\circ \eta_M and the lower arrow (SMgN)Tgη M(S'M\stackrel{g}\to N)\mapsto T'g\circ\eta'_M. Thus the condition renders as

    T(fϕ M)η M=ψ NTfη M T'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M

    or TfTϕ Mη M=Tfψ SMη MT'f\circ T'\phi_M\circ\eta'_M = T'f\circ \psi_{SM}\circ\eta_M. Given ϕ\phi, the uniqueness of B(,ψ)B(-,\psi) is clear from the above diagram, as the horizontal arrows are invertible. B(,ψ)B(-,\psi) determines ψ\psi, namely ψ N=B(,ψ)(id N)\psi_N = B(-,\psi)(id_N). For the existence of ψ\psi (given ϕ\phi) satisfying the above equation, one proposes that ψ\psi is the composition ψ=TεTϕTηT\psi = T'\epsilon \circ T'\phi T \circ \eta'T, i.e.

    TηTTSTTψTTSTTεT 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 μ T\mu^T to δ G\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 ?

    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    • (edited Apr 24th 2011)

    Well now I got it correct in source code. Number 3. Except no formulas in the output!!

    • CommentRowNumber6.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    T(fϕ M)η M=ψ NTfη MT'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M
    • CommentRowNumber7.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    T'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M
    • CommentRowNumber8.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    Strange 6 has a formula in double dollar sign in Markdown+Itex and it does not appear. 7 has the same formula as text output. Over there, the machine has changed the dollar signs into latex in brackets front and back and appeared as a formula. Is somebody experimenting ?
    • CommentRowNumber9.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    • (edited Apr 24th 2011)

    More experimenting

    T(fϕ M)η M=ψ NTfη M T'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M
    • CommentRowNumber10.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011

    Urs’s formula

    Δ n×A n σA nΔ n \Delta^n \times A^n \simeq \coprod_{\sigma \in A^n} \Delta^n
    • CommentRowNumber11.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    T(fϕ M)η M=ψ NTfη M T'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M
    • CommentRowNumber12.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    T(fϕ M)η M=ψ NTfη M T'(f\circ\phi_M)\circ\eta'_M = \psi_N\circ Tf\circ\eta_M
    • CommentRowNumber13.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    Δ n×A n σA nΔ n \Delta^n \times A^n \simeq \coprod_{\sigma \in A^n} \Delta^n
    • CommentRowNumber14.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    T(fϕ M)η M=ψ NTfη M T'(f \circ \phi_M) \circ \eta'_M = \psi_N \circ Tf \circ \eta_M
    • CommentRowNumber15.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    • (edited Apr 24th 2011)
    T(fϕ M)η M T(f \circ \phi_M) \circ \eta_M

    I am not getting it…

    • CommentRowNumber16.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    T(fϕ M)η T'(f \phi_M) \eta
    • CommentRowNumber17.
    • CommentAuthorzskoda
    • CommentTimeApr 24th 2011
    A A
    • CommentRowNumber18.
    • CommentAuthorMike Shulman
    • CommentTimeApr 25th 2011

    I still cannot see the math in any of the above comments.

    • CommentRowNumber19.
    • CommentAuthorzskoda
    • CommentTimeApr 25th 2011
    • (edited Apr 25th 2011)

    Let me try to reinput, that should rerender…I hope.

    TηTTSTTψTTSTTεT T\stackrel{\eta' T}\longrightarrow T'S' T\stackrel{T'\psi T}\longrightarrow T'ST \stackrel{T'\epsilon}\longrightarrow T'
    • CommentRowNumber20.
    • CommentAuthorzskoda
    • CommentTimeApr 25th 2011
    • (edited Apr 25th 2011)

    Yes, finally!

    Edit: well no. I made some changes to 4 and some formulas now appeared but some did not.

    • CommentRowNumber21.
    • CommentAuthorzskoda
    • CommentTimeApr 25th 2011

    It is strange that the math code from my long comment 4 works when copied and pasted into adjoint monad (new entry announced here).

    • CommentRowNumber22.
    • CommentAuthorMike Shulman
    • CommentTimeApr 26th 2011

    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”?

    • CommentRowNumber23.
    • CommentAuthorzskoda
    • CommentTimeApr 26th 2011
    • (edited Apr 26th 2011)

    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 PQP\dashv Q, P:DDP: D'\to D, Q:DDQ:D\to D' and two functors T:DDT:D\to D, T:DDT':D'\to D'. Then there is a correspondence between the transformations ϕ:PTTP\phi:PT'\Rightarrow TP and the transformations ψ:TQQT\psi: T'Q\Rightarrow QT.

    The nnLab entry mate, as I now see, has two adjunctions and two 1-cells to start with. I did not see that general version. Thanks.

    • CommentRowNumber24.
    • CommentAuthorUrs
    • CommentTimeMay 18th 2011

    spelled out at adjoint functor in detail the basic proof that a right adjoint is faithful precisely if the counit has epi components.

    • CommentRowNumber25.
    • CommentAuthoreparejatobes
    • CommentTimeMar 12th 2012

    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

    • CommentRowNumber26.
    • CommentAuthorUrs
    • CommentTimeMar 12th 2012

    Thanks.

    For the sake of those readers who come across this and don’t (yet) know what “LanLan” stands for and who don’t (yet) know what “absolute” refers to, I have slightly edited to read as follows:

    Given L:CDL \colon C \to D, we have that it has a right adjoint R:DCR\colon D \to C precisely if the left Kan extension Lan L1 CLan_L 1_C of the identity along LL exists and is absolute, in which case RLan L1 CR \simeq \mathop{Lan}_L 1_C.

    What do you think?

    • CommentRowNumber27.
    • CommentAuthoreparejatobes
    • CommentTimeMar 12th 2012

    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 \simeqs.

    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

    • CommentRowNumber28.
    • CommentAuthorUrs
    • CommentTimeMar 12th 2012

    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).

    • CommentRowNumber29.
    • CommentAuthoreparejatobes
    • CommentTimeMar 12th 2012

    Ok, many thanks Urs

    • CommentRowNumber30.
    • CommentAuthorMike Shulman
    • CommentTimeMar 12th 2012

    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).

    • CommentRowNumber31.
    • CommentAuthoreparejatobes
    • CommentTimeMar 13th 2012

    done! added a rel link to the absolute kan extension section

    • CommentRowNumber32.
    • CommentAuthorUrs
    • CommentTimeNov 28th 2014

    added more historical references

    • CommentRowNumber33.
    • CommentAuthorUrs
    • CommentTimeJun 2nd 2018

    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.

    diff, v70, current

    • CommentRowNumber34.
    • CommentAuthorUrs
    • CommentTimeJun 2nd 2018

    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.

    • CommentRowNumber35.
    • CommentAuthorUrs
    • CommentTimeJun 3rd 2018
    • (edited Jun 3rd 2018)

    Now I have polished and expanded the next section: In terms of representable functors

    diff, v71, current

    • CommentRowNumber36.
    • CommentAuthorUrs
    • CommentTimeJun 3rd 2018

    now I have similarly touched the next section: In terms of universal factorization through a (co)unit

    diff, v71, current

    • CommentRowNumber37.
    • CommentAuthorUrs
    • CommentTimeJun 3rd 2018

    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)

    diff, v73, current

    • CommentRowNumber38.
    • CommentAuthorUrs
    • CommentTimeJun 20th 2018

    I have polished and expanded the discussion of left adjoints via pointwise limits over comma categories, now this Prop..

    diff, v80, current

    • CommentRowNumber39.
    • CommentAuthorDavid_Corfield
    • CommentTimeSep 2nd 2018

    Corrected the order of adjoint functors in one place.

    diff, v84, current

    • CommentRowNumber40.
    • CommentAuthorUrs
    • CommentTimeJan 28th 2019
    • (edited Jan 28th 2019)

    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]
    

    ?

    diff, v87, current

    • CommentRowNumber41.
    • CommentAuthorMike Shulman
    • CommentTimeJan 28th 2019

    \ar[r,<-] or \ar[from=r] (I’m not actually sure if these is a difference between these)

    • CommentRowNumber42.
    • CommentAuthorUrs
    • CommentTimeJan 29th 2019

    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.

    • CommentRowNumber43.
    • CommentAuthorRichard Williamson
    • CommentTimeJan 29th 2019
    • (edited Jan 29th 2019)

    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.

  1. The issues with escaping should now be fixed.

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