$2$-Morphisms in the Fundamental $2$-Groupoid

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I'm trying to write down a clean definition of the fundamental $2$-groupoid $pi_leq 2(X)$ of a topological space $X$. Specifically, I'm concerned with how to properly define $2$-morphisms. Here is what I have:



The fundamental $2$-groupoid $pi_leq 2(X)$ is the strict $2$-category where



  • The objects of $pi_leq 2(X)$ are the points of $X$.

  • The $1$-morphisms $xrightarrow y$ of $pi_leq 2(X)$ are the
    continuous maps $f:[0,|f|]rightarrow X$ with $f(0)=x$
    and $f(|f|)=y$. Here $|f|geq 0$.

  • For $xxrightarrowfyxrightarrowgz$ define $gcirc f$ as the continuous
    map $[0,|f|+|g|]rightarrow X$ given by
    $$
    (gcirc f)(t)
    =
    begincases
    f(t) & 0leq tleq|f| \
    g(t-|f|) & |f|leq tleq|f|+|g|
    endcases
    $$

Of course, we are deviating from the usual definition of a path as a map $[0,1]rightarrow X$ to make the composition of $1$-morphisms associative on the nose. However, this seems to create a new problem. The $2$-morphisms should be homotopies of paths, but a homotopy between two maps is only defined if the maps have common domain and codomain. To remedy the situation, I've tried the following:



  • For $xxrightarrowfy$, let $overlinef:xrightarrow y$ be the continuous map $[0,1]rightarrow X$ given by $overlinef(t)=f(|f|cdot t)$. Then the $2$-morphisms $alpha:fRightarrow g$ in $pi_leq 2(X)$ are the continuous maps $alpha:[0,1]times[0,1]rightarrow X$ with $alpha(t,0)=overlinef(t)$, $alpha(t,1)=overlineg(t)$, $alpha(0,t)=x$, and $alpha(1,t)=y$.

Is this the standard construction? I'm having trouble finding literature on this.







share|cite|improve this question


















  • 2




    Don't forget that, because you are truncating at $2$, your $2$-morphisms shouldn't be maps, but instead be homotopy equivalence classes of maps.
    – Hurkyl
    Jan 29 '14 at 5:23










  • Oh right. So is my definition correct, up to homotopy?
    – Brian Fitzpatrick
    Jan 29 '14 at 5:25










  • Probably, but I have little experience so I'll leave it for someone else to answer. Incidentally, if I was taking your approach for the $1$-morphisms, I probably would have defined homotopies to be maps from the relevant trapezoids rather than from squares. Of course, that makes composition annoying. :(
    – Hurkyl
    Jan 29 '14 at 5:26







  • 2




    ... and then I'd probably decide that the domains should be a homotopy type, then get annoyed the construction is getting complicated, then go back to squares rather than deal with it. :(
    – Hurkyl
    Jan 29 '14 at 5:30






  • 1




    You meet try working with Moore rectangles: see arXiv:0909.2212 Moore hyperrectangles on a space form a strict cubical omega-category. But in the context of spaces, and no extra structure, there will always be problems with getting a strict 2-groupoid or double groupoid, in my view.
    – Ronnie Brown
    Feb 2 '14 at 23:10














up vote
6
down vote

favorite
1












I'm trying to write down a clean definition of the fundamental $2$-groupoid $pi_leq 2(X)$ of a topological space $X$. Specifically, I'm concerned with how to properly define $2$-morphisms. Here is what I have:



The fundamental $2$-groupoid $pi_leq 2(X)$ is the strict $2$-category where



  • The objects of $pi_leq 2(X)$ are the points of $X$.

  • The $1$-morphisms $xrightarrow y$ of $pi_leq 2(X)$ are the
    continuous maps $f:[0,|f|]rightarrow X$ with $f(0)=x$
    and $f(|f|)=y$. Here $|f|geq 0$.

  • For $xxrightarrowfyxrightarrowgz$ define $gcirc f$ as the continuous
    map $[0,|f|+|g|]rightarrow X$ given by
    $$
    (gcirc f)(t)
    =
    begincases
    f(t) & 0leq tleq|f| \
    g(t-|f|) & |f|leq tleq|f|+|g|
    endcases
    $$

Of course, we are deviating from the usual definition of a path as a map $[0,1]rightarrow X$ to make the composition of $1$-morphisms associative on the nose. However, this seems to create a new problem. The $2$-morphisms should be homotopies of paths, but a homotopy between two maps is only defined if the maps have common domain and codomain. To remedy the situation, I've tried the following:



  • For $xxrightarrowfy$, let $overlinef:xrightarrow y$ be the continuous map $[0,1]rightarrow X$ given by $overlinef(t)=f(|f|cdot t)$. Then the $2$-morphisms $alpha:fRightarrow g$ in $pi_leq 2(X)$ are the continuous maps $alpha:[0,1]times[0,1]rightarrow X$ with $alpha(t,0)=overlinef(t)$, $alpha(t,1)=overlineg(t)$, $alpha(0,t)=x$, and $alpha(1,t)=y$.

Is this the standard construction? I'm having trouble finding literature on this.







share|cite|improve this question


















  • 2




    Don't forget that, because you are truncating at $2$, your $2$-morphisms shouldn't be maps, but instead be homotopy equivalence classes of maps.
    – Hurkyl
    Jan 29 '14 at 5:23










  • Oh right. So is my definition correct, up to homotopy?
    – Brian Fitzpatrick
    Jan 29 '14 at 5:25










  • Probably, but I have little experience so I'll leave it for someone else to answer. Incidentally, if I was taking your approach for the $1$-morphisms, I probably would have defined homotopies to be maps from the relevant trapezoids rather than from squares. Of course, that makes composition annoying. :(
    – Hurkyl
    Jan 29 '14 at 5:26







  • 2




    ... and then I'd probably decide that the domains should be a homotopy type, then get annoyed the construction is getting complicated, then go back to squares rather than deal with it. :(
    – Hurkyl
    Jan 29 '14 at 5:30






  • 1




    You meet try working with Moore rectangles: see arXiv:0909.2212 Moore hyperrectangles on a space form a strict cubical omega-category. But in the context of spaces, and no extra structure, there will always be problems with getting a strict 2-groupoid or double groupoid, in my view.
    – Ronnie Brown
    Feb 2 '14 at 23:10












up vote
6
down vote

favorite
1









up vote
6
down vote

favorite
1






1





I'm trying to write down a clean definition of the fundamental $2$-groupoid $pi_leq 2(X)$ of a topological space $X$. Specifically, I'm concerned with how to properly define $2$-morphisms. Here is what I have:



The fundamental $2$-groupoid $pi_leq 2(X)$ is the strict $2$-category where



  • The objects of $pi_leq 2(X)$ are the points of $X$.

  • The $1$-morphisms $xrightarrow y$ of $pi_leq 2(X)$ are the
    continuous maps $f:[0,|f|]rightarrow X$ with $f(0)=x$
    and $f(|f|)=y$. Here $|f|geq 0$.

  • For $xxrightarrowfyxrightarrowgz$ define $gcirc f$ as the continuous
    map $[0,|f|+|g|]rightarrow X$ given by
    $$
    (gcirc f)(t)
    =
    begincases
    f(t) & 0leq tleq|f| \
    g(t-|f|) & |f|leq tleq|f|+|g|
    endcases
    $$

Of course, we are deviating from the usual definition of a path as a map $[0,1]rightarrow X$ to make the composition of $1$-morphisms associative on the nose. However, this seems to create a new problem. The $2$-morphisms should be homotopies of paths, but a homotopy between two maps is only defined if the maps have common domain and codomain. To remedy the situation, I've tried the following:



  • For $xxrightarrowfy$, let $overlinef:xrightarrow y$ be the continuous map $[0,1]rightarrow X$ given by $overlinef(t)=f(|f|cdot t)$. Then the $2$-morphisms $alpha:fRightarrow g$ in $pi_leq 2(X)$ are the continuous maps $alpha:[0,1]times[0,1]rightarrow X$ with $alpha(t,0)=overlinef(t)$, $alpha(t,1)=overlineg(t)$, $alpha(0,t)=x$, and $alpha(1,t)=y$.

Is this the standard construction? I'm having trouble finding literature on this.







share|cite|improve this question














I'm trying to write down a clean definition of the fundamental $2$-groupoid $pi_leq 2(X)$ of a topological space $X$. Specifically, I'm concerned with how to properly define $2$-morphisms. Here is what I have:



The fundamental $2$-groupoid $pi_leq 2(X)$ is the strict $2$-category where



  • The objects of $pi_leq 2(X)$ are the points of $X$.

  • The $1$-morphisms $xrightarrow y$ of $pi_leq 2(X)$ are the
    continuous maps $f:[0,|f|]rightarrow X$ with $f(0)=x$
    and $f(|f|)=y$. Here $|f|geq 0$.

  • For $xxrightarrowfyxrightarrowgz$ define $gcirc f$ as the continuous
    map $[0,|f|+|g|]rightarrow X$ given by
    $$
    (gcirc f)(t)
    =
    begincases
    f(t) & 0leq tleq|f| \
    g(t-|f|) & |f|leq tleq|f|+|g|
    endcases
    $$

Of course, we are deviating from the usual definition of a path as a map $[0,1]rightarrow X$ to make the composition of $1$-morphisms associative on the nose. However, this seems to create a new problem. The $2$-morphisms should be homotopies of paths, but a homotopy between two maps is only defined if the maps have common domain and codomain. To remedy the situation, I've tried the following:



  • For $xxrightarrowfy$, let $overlinef:xrightarrow y$ be the continuous map $[0,1]rightarrow X$ given by $overlinef(t)=f(|f|cdot t)$. Then the $2$-morphisms $alpha:fRightarrow g$ in $pi_leq 2(X)$ are the continuous maps $alpha:[0,1]times[0,1]rightarrow X$ with $alpha(t,0)=overlinef(t)$, $alpha(t,1)=overlineg(t)$, $alpha(0,t)=x$, and $alpha(1,t)=y$.

Is this the standard construction? I'm having trouble finding literature on this.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 '14 at 4:43

























asked Jan 29 '14 at 4:36









Brian Fitzpatrick

20.7k42958




20.7k42958







  • 2




    Don't forget that, because you are truncating at $2$, your $2$-morphisms shouldn't be maps, but instead be homotopy equivalence classes of maps.
    – Hurkyl
    Jan 29 '14 at 5:23










  • Oh right. So is my definition correct, up to homotopy?
    – Brian Fitzpatrick
    Jan 29 '14 at 5:25










  • Probably, but I have little experience so I'll leave it for someone else to answer. Incidentally, if I was taking your approach for the $1$-morphisms, I probably would have defined homotopies to be maps from the relevant trapezoids rather than from squares. Of course, that makes composition annoying. :(
    – Hurkyl
    Jan 29 '14 at 5:26







  • 2




    ... and then I'd probably decide that the domains should be a homotopy type, then get annoyed the construction is getting complicated, then go back to squares rather than deal with it. :(
    – Hurkyl
    Jan 29 '14 at 5:30






  • 1




    You meet try working with Moore rectangles: see arXiv:0909.2212 Moore hyperrectangles on a space form a strict cubical omega-category. But in the context of spaces, and no extra structure, there will always be problems with getting a strict 2-groupoid or double groupoid, in my view.
    – Ronnie Brown
    Feb 2 '14 at 23:10












  • 2




    Don't forget that, because you are truncating at $2$, your $2$-morphisms shouldn't be maps, but instead be homotopy equivalence classes of maps.
    – Hurkyl
    Jan 29 '14 at 5:23










  • Oh right. So is my definition correct, up to homotopy?
    – Brian Fitzpatrick
    Jan 29 '14 at 5:25










  • Probably, but I have little experience so I'll leave it for someone else to answer. Incidentally, if I was taking your approach for the $1$-morphisms, I probably would have defined homotopies to be maps from the relevant trapezoids rather than from squares. Of course, that makes composition annoying. :(
    – Hurkyl
    Jan 29 '14 at 5:26







  • 2




    ... and then I'd probably decide that the domains should be a homotopy type, then get annoyed the construction is getting complicated, then go back to squares rather than deal with it. :(
    – Hurkyl
    Jan 29 '14 at 5:30






  • 1




    You meet try working with Moore rectangles: see arXiv:0909.2212 Moore hyperrectangles on a space form a strict cubical omega-category. But in the context of spaces, and no extra structure, there will always be problems with getting a strict 2-groupoid or double groupoid, in my view.
    – Ronnie Brown
    Feb 2 '14 at 23:10







2




2




Don't forget that, because you are truncating at $2$, your $2$-morphisms shouldn't be maps, but instead be homotopy equivalence classes of maps.
– Hurkyl
Jan 29 '14 at 5:23




Don't forget that, because you are truncating at $2$, your $2$-morphisms shouldn't be maps, but instead be homotopy equivalence classes of maps.
– Hurkyl
Jan 29 '14 at 5:23












Oh right. So is my definition correct, up to homotopy?
– Brian Fitzpatrick
Jan 29 '14 at 5:25




Oh right. So is my definition correct, up to homotopy?
– Brian Fitzpatrick
Jan 29 '14 at 5:25












Probably, but I have little experience so I'll leave it for someone else to answer. Incidentally, if I was taking your approach for the $1$-morphisms, I probably would have defined homotopies to be maps from the relevant trapezoids rather than from squares. Of course, that makes composition annoying. :(
– Hurkyl
Jan 29 '14 at 5:26





Probably, but I have little experience so I'll leave it for someone else to answer. Incidentally, if I was taking your approach for the $1$-morphisms, I probably would have defined homotopies to be maps from the relevant trapezoids rather than from squares. Of course, that makes composition annoying. :(
– Hurkyl
Jan 29 '14 at 5:26





2




2




... and then I'd probably decide that the domains should be a homotopy type, then get annoyed the construction is getting complicated, then go back to squares rather than deal with it. :(
– Hurkyl
Jan 29 '14 at 5:30




... and then I'd probably decide that the domains should be a homotopy type, then get annoyed the construction is getting complicated, then go back to squares rather than deal with it. :(
– Hurkyl
Jan 29 '14 at 5:30




1




1




You meet try working with Moore rectangles: see arXiv:0909.2212 Moore hyperrectangles on a space form a strict cubical omega-category. But in the context of spaces, and no extra structure, there will always be problems with getting a strict 2-groupoid or double groupoid, in my view.
– Ronnie Brown
Feb 2 '14 at 23:10




You meet try working with Moore rectangles: see arXiv:0909.2212 Moore hyperrectangles on a space form a strict cubical omega-category. But in the context of spaces, and no extra structure, there will always be problems with getting a strict 2-groupoid or double groupoid, in my view.
– Ronnie Brown
Feb 2 '14 at 23:10










1 Answer
1






active

oldest

votes

















up vote
4
down vote













This is a question which raises all sorts of interesting issues!



I think it is easier to look for double groupoids rather than $2$-groupoids, because it is more symmetric with regard to directions and before taking homotopy classes of some kind, the notion of partial composition in a given direction is very clear, and extends to all dimensions. See also this question on mathoverflow.



I though about this for 9 years starting in 1965 with the aim, as it seemed likely that the proof I had written out many times for my topology book extended to get a $2$-dimensional van Kampen type theorem, and got myself quite confused. So it was "an idea for a proof in search of a theorem".



The breakthrough came with Philip Higgins in 1974 when, as suggested by work of J H C Whitehead on second relative homotopy groups, we started looking not at spaces but at pairs of spaces, $(X,A)$. So the natural idea was to look at maps of square $I^2$ into $X$ which took the vertices to the base point and edges into $A$. Then all became easy, not trivial, but easy. The details are in this book, which has a free pdf. A key picture for the definition of compositions being well defined is



pic



The point is that that if $alpha, beta$ are two squares as above then their classes $[alpha], [beta]$ are homotopy classes rel vertices of maps $I^2 to X$ where during the homotopies the boundary of $I^2$ stays in $A$. So $ partial^+_2[alpha] =partial^2_-[beta] $ means there is a homotopy $h$ of two loops in $A$, and such a homotopy is arbitrary. So when you try to prove independence of choices, you get the above figure, which has a big hole in the middle! But the bottom edges of the hole are all constant, so you can fill in the bottom square. Now fill in the hole by a retraction. All the faces of the "hole" are in $A$ so in the retraction the top face is also in $A$. So the whole block gives a homotopy $$alpha +_2 h +_2 beta simeq alpha' +_2 h'+_2 beta'$$ of the type required.



This construction we called $rho_2(X,A,a)$. It contains $pi_2(X,A,a)$ . From this you can get as a substructure a $2$-groupoid, if you really want it.



There are papers dealing with the absolute case for Hausdorff spaces, see K.A. Hardie, K.H. Kamps and R.W. Kieboom, ‘A homotopy 2–groupoid of a Hausdorff space’, Applied Cat. Structures 8 (2000), 209–234, and subsequent papers dealing with double groupoids in a related fashion. But these papers have not it seems led to new calculations of homotopical invariants, as has the above relative construction.



Edit Feb 9, 2014: The above construction of a homotopy double groupoid of a pair is used to prove a $2$-dimensional van Kampen type theorem, one of whose consequences, an excision type theorem for second relative homotopy groups, is explained in my answer to this stackexchange question.






share|cite|improve this answer






















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    1 Answer
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    active

    oldest

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    active

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    up vote
    4
    down vote













    This is a question which raises all sorts of interesting issues!



    I think it is easier to look for double groupoids rather than $2$-groupoids, because it is more symmetric with regard to directions and before taking homotopy classes of some kind, the notion of partial composition in a given direction is very clear, and extends to all dimensions. See also this question on mathoverflow.



    I though about this for 9 years starting in 1965 with the aim, as it seemed likely that the proof I had written out many times for my topology book extended to get a $2$-dimensional van Kampen type theorem, and got myself quite confused. So it was "an idea for a proof in search of a theorem".



    The breakthrough came with Philip Higgins in 1974 when, as suggested by work of J H C Whitehead on second relative homotopy groups, we started looking not at spaces but at pairs of spaces, $(X,A)$. So the natural idea was to look at maps of square $I^2$ into $X$ which took the vertices to the base point and edges into $A$. Then all became easy, not trivial, but easy. The details are in this book, which has a free pdf. A key picture for the definition of compositions being well defined is



    pic



    The point is that that if $alpha, beta$ are two squares as above then their classes $[alpha], [beta]$ are homotopy classes rel vertices of maps $I^2 to X$ where during the homotopies the boundary of $I^2$ stays in $A$. So $ partial^+_2[alpha] =partial^2_-[beta] $ means there is a homotopy $h$ of two loops in $A$, and such a homotopy is arbitrary. So when you try to prove independence of choices, you get the above figure, which has a big hole in the middle! But the bottom edges of the hole are all constant, so you can fill in the bottom square. Now fill in the hole by a retraction. All the faces of the "hole" are in $A$ so in the retraction the top face is also in $A$. So the whole block gives a homotopy $$alpha +_2 h +_2 beta simeq alpha' +_2 h'+_2 beta'$$ of the type required.



    This construction we called $rho_2(X,A,a)$. It contains $pi_2(X,A,a)$ . From this you can get as a substructure a $2$-groupoid, if you really want it.



    There are papers dealing with the absolute case for Hausdorff spaces, see K.A. Hardie, K.H. Kamps and R.W. Kieboom, ‘A homotopy 2–groupoid of a Hausdorff space’, Applied Cat. Structures 8 (2000), 209–234, and subsequent papers dealing with double groupoids in a related fashion. But these papers have not it seems led to new calculations of homotopical invariants, as has the above relative construction.



    Edit Feb 9, 2014: The above construction of a homotopy double groupoid of a pair is used to prove a $2$-dimensional van Kampen type theorem, one of whose consequences, an excision type theorem for second relative homotopy groups, is explained in my answer to this stackexchange question.






    share|cite|improve this answer


























      up vote
      4
      down vote













      This is a question which raises all sorts of interesting issues!



      I think it is easier to look for double groupoids rather than $2$-groupoids, because it is more symmetric with regard to directions and before taking homotopy classes of some kind, the notion of partial composition in a given direction is very clear, and extends to all dimensions. See also this question on mathoverflow.



      I though about this for 9 years starting in 1965 with the aim, as it seemed likely that the proof I had written out many times for my topology book extended to get a $2$-dimensional van Kampen type theorem, and got myself quite confused. So it was "an idea for a proof in search of a theorem".



      The breakthrough came with Philip Higgins in 1974 when, as suggested by work of J H C Whitehead on second relative homotopy groups, we started looking not at spaces but at pairs of spaces, $(X,A)$. So the natural idea was to look at maps of square $I^2$ into $X$ which took the vertices to the base point and edges into $A$. Then all became easy, not trivial, but easy. The details are in this book, which has a free pdf. A key picture for the definition of compositions being well defined is



      pic



      The point is that that if $alpha, beta$ are two squares as above then their classes $[alpha], [beta]$ are homotopy classes rel vertices of maps $I^2 to X$ where during the homotopies the boundary of $I^2$ stays in $A$. So $ partial^+_2[alpha] =partial^2_-[beta] $ means there is a homotopy $h$ of two loops in $A$, and such a homotopy is arbitrary. So when you try to prove independence of choices, you get the above figure, which has a big hole in the middle! But the bottom edges of the hole are all constant, so you can fill in the bottom square. Now fill in the hole by a retraction. All the faces of the "hole" are in $A$ so in the retraction the top face is also in $A$. So the whole block gives a homotopy $$alpha +_2 h +_2 beta simeq alpha' +_2 h'+_2 beta'$$ of the type required.



      This construction we called $rho_2(X,A,a)$. It contains $pi_2(X,A,a)$ . From this you can get as a substructure a $2$-groupoid, if you really want it.



      There are papers dealing with the absolute case for Hausdorff spaces, see K.A. Hardie, K.H. Kamps and R.W. Kieboom, ‘A homotopy 2–groupoid of a Hausdorff space’, Applied Cat. Structures 8 (2000), 209–234, and subsequent papers dealing with double groupoids in a related fashion. But these papers have not it seems led to new calculations of homotopical invariants, as has the above relative construction.



      Edit Feb 9, 2014: The above construction of a homotopy double groupoid of a pair is used to prove a $2$-dimensional van Kampen type theorem, one of whose consequences, an excision type theorem for second relative homotopy groups, is explained in my answer to this stackexchange question.






      share|cite|improve this answer
























        up vote
        4
        down vote










        up vote
        4
        down vote









        This is a question which raises all sorts of interesting issues!



        I think it is easier to look for double groupoids rather than $2$-groupoids, because it is more symmetric with regard to directions and before taking homotopy classes of some kind, the notion of partial composition in a given direction is very clear, and extends to all dimensions. See also this question on mathoverflow.



        I though about this for 9 years starting in 1965 with the aim, as it seemed likely that the proof I had written out many times for my topology book extended to get a $2$-dimensional van Kampen type theorem, and got myself quite confused. So it was "an idea for a proof in search of a theorem".



        The breakthrough came with Philip Higgins in 1974 when, as suggested by work of J H C Whitehead on second relative homotopy groups, we started looking not at spaces but at pairs of spaces, $(X,A)$. So the natural idea was to look at maps of square $I^2$ into $X$ which took the vertices to the base point and edges into $A$. Then all became easy, not trivial, but easy. The details are in this book, which has a free pdf. A key picture for the definition of compositions being well defined is



        pic



        The point is that that if $alpha, beta$ are two squares as above then their classes $[alpha], [beta]$ are homotopy classes rel vertices of maps $I^2 to X$ where during the homotopies the boundary of $I^2$ stays in $A$. So $ partial^+_2[alpha] =partial^2_-[beta] $ means there is a homotopy $h$ of two loops in $A$, and such a homotopy is arbitrary. So when you try to prove independence of choices, you get the above figure, which has a big hole in the middle! But the bottom edges of the hole are all constant, so you can fill in the bottom square. Now fill in the hole by a retraction. All the faces of the "hole" are in $A$ so in the retraction the top face is also in $A$. So the whole block gives a homotopy $$alpha +_2 h +_2 beta simeq alpha' +_2 h'+_2 beta'$$ of the type required.



        This construction we called $rho_2(X,A,a)$. It contains $pi_2(X,A,a)$ . From this you can get as a substructure a $2$-groupoid, if you really want it.



        There are papers dealing with the absolute case for Hausdorff spaces, see K.A. Hardie, K.H. Kamps and R.W. Kieboom, ‘A homotopy 2–groupoid of a Hausdorff space’, Applied Cat. Structures 8 (2000), 209–234, and subsequent papers dealing with double groupoids in a related fashion. But these papers have not it seems led to new calculations of homotopical invariants, as has the above relative construction.



        Edit Feb 9, 2014: The above construction of a homotopy double groupoid of a pair is used to prove a $2$-dimensional van Kampen type theorem, one of whose consequences, an excision type theorem for second relative homotopy groups, is explained in my answer to this stackexchange question.






        share|cite|improve this answer














        This is a question which raises all sorts of interesting issues!



        I think it is easier to look for double groupoids rather than $2$-groupoids, because it is more symmetric with regard to directions and before taking homotopy classes of some kind, the notion of partial composition in a given direction is very clear, and extends to all dimensions. See also this question on mathoverflow.



        I though about this for 9 years starting in 1965 with the aim, as it seemed likely that the proof I had written out many times for my topology book extended to get a $2$-dimensional van Kampen type theorem, and got myself quite confused. So it was "an idea for a proof in search of a theorem".



        The breakthrough came with Philip Higgins in 1974 when, as suggested by work of J H C Whitehead on second relative homotopy groups, we started looking not at spaces but at pairs of spaces, $(X,A)$. So the natural idea was to look at maps of square $I^2$ into $X$ which took the vertices to the base point and edges into $A$. Then all became easy, not trivial, but easy. The details are in this book, which has a free pdf. A key picture for the definition of compositions being well defined is



        pic



        The point is that that if $alpha, beta$ are two squares as above then their classes $[alpha], [beta]$ are homotopy classes rel vertices of maps $I^2 to X$ where during the homotopies the boundary of $I^2$ stays in $A$. So $ partial^+_2[alpha] =partial^2_-[beta] $ means there is a homotopy $h$ of two loops in $A$, and such a homotopy is arbitrary. So when you try to prove independence of choices, you get the above figure, which has a big hole in the middle! But the bottom edges of the hole are all constant, so you can fill in the bottom square. Now fill in the hole by a retraction. All the faces of the "hole" are in $A$ so in the retraction the top face is also in $A$. So the whole block gives a homotopy $$alpha +_2 h +_2 beta simeq alpha' +_2 h'+_2 beta'$$ of the type required.



        This construction we called $rho_2(X,A,a)$. It contains $pi_2(X,A,a)$ . From this you can get as a substructure a $2$-groupoid, if you really want it.



        There are papers dealing with the absolute case for Hausdorff spaces, see K.A. Hardie, K.H. Kamps and R.W. Kieboom, ‘A homotopy 2–groupoid of a Hausdorff space’, Applied Cat. Structures 8 (2000), 209–234, and subsequent papers dealing with double groupoids in a related fashion. But these papers have not it seems led to new calculations of homotopical invariants, as has the above relative construction.



        Edit Feb 9, 2014: The above construction of a homotopy double groupoid of a pair is used to prove a $2$-dimensional van Kampen type theorem, one of whose consequences, an excision type theorem for second relative homotopy groups, is explained in my answer to this stackexchange question.







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        edited Aug 16 at 10:25

























        answered Jan 31 '14 at 18:52









        Ronnie Brown

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