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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeMay 29th 2010
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   Author: Mike Shulman
   Format: MarkdownItexCreated [[exact square]], but haven't linked to it from anywhere else yet. I'm planning to move some of the discussion of exactness at [[derivator]] to its own page [[homotopy exact square]], analogous to this one. (Is the phrase "exact square" used for other things that we should worry about disambiguating/clarifying?)

   Created exact square, but haven’t linked to it from anywhere else yet. I’m planning to move some of the discussion of exactness at derivator to its own page homotopy exact square, analogous to this one.

   (Is the phrase “exact square” used for other things that we should worry about disambiguating/clarifying?)

2. • CommentRowNumber2.
   • CommentAuthorTim_Porter
   • CommentTimeMay 30th 2010
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   Author: Tim_Porter
   Format: MarkdownItexA bit of History: The original idea of exact square (as so named) is I think in some work by Hilton in an Abelian category context. Guitart then studied them in a series of articles starting in 1980, and a discussion of various aspects of their use can be found in Cordier-Porter: Shape Theory, Dover 2008. I do not think there is another older source for the material than Guitart's papers and no other concept is called that as far as I remember.

   A bit of History: The original idea of exact square (as so named) is I think in some work by Hilton in an Abelian category context. Guitart then studied them in a series of articles starting in 1980, and a discussion of various aspects of their use can be found in Cordier-Porter: Shape Theory, Dover 2008. I do not think there is another older source for the material than Guitart’s papers and no other concept is called that as far as I remember.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeMay 30th 2010
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   Author: Mike Shulman
   Format: MarkdownItexInteresting, thanks; why not add some of that to the page? (-:

   Interesting, thanks; why not add some of that to the page? (-:

4. • CommentRowNumber4.
   • CommentAuthorTim_Porter
   • CommentTimeMay 31st 2010
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   Author: Tim_Porter
   Format: MarkdownItexI will when I get time! :-) The page is pretty good already. I have a plan to write up some older stuff on categorical shape theory including exact squares, then Batanin's strong shape version of it and then to look at it from a nPOV. That is one plan. (There are others :-( !!!)

   I will when I get time! :-) The page is pretty good already. I have a plan to write up some older stuff on categorical shape theory including exact squares, then Batanin’s strong shape version of it and then to look at it from a nPOV. That is one plan. (There are others :-( !!!)

5. • CommentRowNumber5.
   • CommentAuthorHarry Gindi
   • CommentTimeMay 31st 2010
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   Author: Harry Gindi
   Format: MarkdownItexSorry to be a bore, but what exactly is the natural transformation between? I assume it's between the compositions of functors, but the way the diagram is drawn is misleading, and the wording is abstruse (inhabited by a natural transformation?).

   Sorry to be a bore, but what exactly is the natural transformation between? I assume it’s between the compositions of functors, but the way the diagram is drawn is misleading, and the wording is abstruse (inhabited by a natural transformation?).

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 31st 2010
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   Author: Todd_Trimble
   Format: MarkdownItex"Inhabited" = filled-in. Seems okay to me. I agree that the way the 2-cell is drawn makes the reader slow down for two seconds to verify there is no ambiguity (there isn't: there's only one way that diagram could be interpreted), and that slow-down isn't necessary. I'm going to fix that.

   “Inhabited” = filled-in. Seems okay to me.

   I agree that the way the 2-cell is drawn makes the reader slow down for two seconds to verify there is no ambiguity (there isn’t: there’s only one way that diagram could be interpreted), and that slow-down isn’t necessary. I’m going to fix that.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeMay 31st 2010
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   Author: Mike Shulman
   Format: MarkdownItexFor some reason I didn't know that Instiki would accept `\swArrow`. I'll use that from now on; it's definitely better to have the arrow going in that direction. (Although I think drawing the arrow as a $\Downarrow$ the way I did originally is fairly common, especially in Australia, so it's not a bad thing to get used to being able to parse when reading.)

   For some reason I didn’t know that Instiki would accept \swArrow. I’ll use that from now on; it’s definitely better to have the arrow going in that direction. (Although I think drawing the arrow as a $\Downarrow$ the way I did originally is fairly common, especially in Australia, so it’s not a bad thing to get used to being able to parse when reading.)

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeJun 8th 2010
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   Author: Mike Shulman
   Format: MarkdownItexThe reason I didn't know about `\swArrow` is that `\swArrow` doesn't exist in my copy of the Comprehensive LaTeX Symbols List! There are a couple of packages listed with `\Swarrow`, however, so I think I assumed that would be the command, tried it and it didn't work, and assumed that itex didn't know how to make a $\swArrow$. Is `\swArrow` taken from any actual LaTeX package? The capitalization seems to clash with `\Rightarrow` etc.

   The reason I didn’t know about \swArrow is that \swArrow doesn’t exist in my copy of the Comprehensive LaTeX Symbols List! There are a couple of packages listed with \Swarrow, however, so I think I assumed that would be the command, tried it and it didn’t work, and assumed that itex didn’t know how to make a $\swArrow$. Is \swArrow taken from any actual LaTeX package? The capitalization seems to clash with \Rightarrow etc.

9. • CommentRowNumber9.
   • CommentAuthorTim_Porter
   • CommentTimeNov 22nd 2011
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   Author: Tim_Porter
   Format: MarkdownItexTom Hirschowitz has put a query on [[exact square]]. I will suggest that he copies it here as his office is only on the next floor downfrom where I am!

   Tom Hirschowitz has put a query on exact square. I will suggest that he copies it here as his office is only on the next floor downfrom where I am!

10. • CommentRowNumber10.
    • CommentAuthorGuest
    • CommentTimeNov 22nd 2011
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    Author: Guest
    Format: MarkdownItexHi, this is Tom doing as told, as Guest waiting for my account to be validated. Here is the query. The criterion for exactness in the set-based setting seems wrong, one symptom being that it does not mention the 2-cell $v f \to g u$ at all. If so, a few things below are probably also wrong. Here is a tentative corrected version (the math don't look good, but I prefer to let them unchanged, as I'm copying them from the nlab query): [...] the above condition means that, calling $\psi \colon g u \to v f$ the given 2-cell, we have 1. For any morphism $\varphi\colon v(b) \to g(c)$ in $D$, there exists an $a\in A$ and morphisms $\alpha\colon u(a) \to c$ and $\beta\colon b\to f(a)$ such that $g(\alpha) \circ \psi_a \circ v(\beta) = \varphi$, and 2. For any $(a,\alpha,\beta)$ and $(a',\alpha',\beta')$ as above with $g(\alpha) \circ \psi_a \circ v(\beta) = g(\alpha') \circ \psi_{a'} \circ v(\beta')$, there is a [[zigzag]] of arrows connecting $a$ to $a'$ and rendering the evident induced diagram commutative.

    Hi, this is Tom doing as told, as Guest waiting for my account to be validated. Here is the query.

    The criterion for exactness in the set-based setting seems wrong, one symptom being that it does not mention the 2-cell $v f \to g u$ at all. If so, a few things below are probably also wrong. Here is a tentative corrected version (the math don’t look good, but I prefer to let them unchanged, as I’m copying them from the nlab query):

    […] the above condition means that, calling $\psi \colon g u \to v f$ the given 2-cell, we have

    1. For any morphism $\varphi\colon v(b) \to g(c)$ in $D$, there exists an $a\in A$ and morphisms $\alpha\colon u(a) \to c$ and $\beta\colon b\to f(a)$ such that $g(\alpha) \circ \psi_a \circ v(\beta) = \varphi$, and

    2. For any $(a,\alpha,\beta)$ and $(a',\alpha',\beta')$ as above with $g(\alpha) \circ \psi_a \circ v(\beta) = g(\alpha') \circ \psi_{a'} \circ v(\beta')$, there is a zigzag of arrows connecting $a$ to $a'$ and rendering the evident induced diagram commutative.

11. • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeNov 22nd 2011
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    Author: Mike Shulman
    Format: MarkdownItexHi Tom, welcome to the nForum! I think you're absolutely right; that explicit version seems to only be correct in the special case when $\psi$ is an identity. Your version looks fine to me; want to fix the page?

    Hi Tom, welcome to the nForum! I think you’re absolutely right; that explicit version seems to only be correct in the special case when $\psi$ is an identity. Your version looks fine to me; want to fix the page?

12. • CommentRowNumber12.
    • CommentAuthorTom Hirschowitz
    • CommentTimeNov 25th 2011
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    Author: Tom Hirschowitz
    Format: TextYes, I'll try.

    Yes, I'll try.
13. • CommentRowNumber13.
    • CommentAuthorTom Hirschowitz
    • CommentTimeSep 18th 2013
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    Author: Tom Hirschowitz
    Format: MarkdownItexHi. Me again, after a few years. I haven't yet tried very hard, but I can't see why the exactness condition is equivalent to $v^* g_* \to f_* u^*$ being an iso. Would anyone have a reference or a hint, please?

    Hi. Me again, after a few years.

    I haven’t yet tried very hard, but I can’t see why the exactness condition is equivalent to $v^* g_* \to f_* u^*$ being an iso. Would anyone have a reference or a hint, please?

14. • CommentRowNumber14.
    • CommentAuthorZhen Lin
    • CommentTimeSep 18th 2013
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    Author: Zhen Lin
    Format: MarkdownItexI assume you are taking $u_! f^* \Rightarrow g^* v_!$ being an isomorphism as your definition. In that case, the two conditions are equivalent because adjunctions can be composed and adjoints are unique up to unique isomorphism: so $g^* v_! \dashv v^* g_*$ and $u_! f^* \dashv f_* u^*$. (The only thing that really needs verifying is that the "unique isomorphism" induced by $u_! f^* \Rightarrow g^* v_!$ is indeed the given $v^* g_* \Rightarrow f_* u^*$, but this can be checked using string diagrams or similar.)

    I assume you are taking $u_! f^* \Rightarrow g^* v_!$ being an isomorphism as your definition. In that case, the two conditions are equivalent because adjunctions can be composed and adjoints are unique up to unique isomorphism: so $g^* v_! \dashv v^* g_*$ and $u_! f^* \dashv f_* u^*$. (The only thing that really needs verifying is that the “unique isomorphism” induced by $u_! f^* \Rightarrow g^* v_!$ is indeed the given $v^* g_* \Rightarrow f_* u^*$, but this can be checked using string diagrams or similar.)

15. • CommentRowNumber15.
    • CommentAuthorTom Hirschowitz
    • CommentTimeSep 19th 2013
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    Author: Tom Hirschowitz
    Format: MarkdownItexAh, right, thanks. Actually this seems to work by pure string diagram calculations.

    Ah, right, thanks. Actually this seems to work by pure string diagram calculations.

16. • CommentRowNumber16.
    • CommentAuthorTim Campion
    • CommentTimeJul 23rd 2016
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    Author: Tim Campion
    Format: MarkdownItexAdded an exact square characterization of adjointness. Also clarified that the exact square characterization of fully faithful functors is an if-and-only-if.

    Added an exact square characterization of adjointness. Also clarified that the exact square characterization of fully faithful functors is an if-and-only-if.