$2$-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.
algebraic-topology category-theory higher-category-theory
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up vote
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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.
algebraic-topology category-theory higher-category-theory
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.
â Hurkyl
Jan 29 '14 at 5:23
Oh right. So is my definition correct, up to homotopy?
â 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. :(
â 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. :(
â 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.
â Ronnie Brown
Feb 2 '14 at 23:10
 |Â
show 1 more comment
up vote
6
down vote
favorite
up vote
6
down vote
favorite
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.
algebraic-topology category-theory higher-category-theory
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.
algebraic-topology category-theory higher-category-theory
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.
â Hurkyl
Jan 29 '14 at 5:23
Oh right. So is my definition correct, up to homotopy?
â 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. :(
â 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. :(
â 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.
â Ronnie Brown
Feb 2 '14 at 23:10
 |Â
show 1 more comment
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.
â Hurkyl
Jan 29 '14 at 5:23
Oh right. So is my definition correct, up to homotopy?
â 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. :(
â 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. :(
â 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.
â 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.
â 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.
â Hurkyl
Jan 29 '14 at 5:23
Oh right. So is my definition correct, up to homotopy?
â Brian Fitzpatrick
Jan 29 '14 at 5:25
Oh right. So is my definition correct, up to homotopy?
â 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. :(
â 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. :(
â 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. :(
â 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. :(
â 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.
â 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.
â Ronnie Brown
Feb 2 '14 at 23:10
 |Â
show 1 more comment
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
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, âÂÂA homotopy 2âÂÂgroupoid of a HausdorﬠspaceâÂÂ, Applied Cat. Structures 8 (2000), 209âÂÂ234, 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.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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
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, âÂÂA homotopy 2âÂÂgroupoid of a HausdorﬠspaceâÂÂ, Applied Cat. Structures 8 (2000), 209âÂÂ234, 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.
add a comment |Â
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
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, âÂÂA homotopy 2âÂÂgroupoid of a HausdorﬠspaceâÂÂ, Applied Cat. Structures 8 (2000), 209âÂÂ234, 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.
add a comment |Â
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
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, âÂÂA homotopy 2âÂÂgroupoid of a HausdorﬠspaceâÂÂ, Applied Cat. Structures 8 (2000), 209âÂÂ234, 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.
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
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, âÂÂA homotopy 2âÂÂgroupoid of a HausdorﬠspaceâÂÂ, Applied Cat. Structures 8 (2000), 209âÂÂ234, 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.
edited Aug 16 at 10:25
answered Jan 31 '14 at 18:52
Ronnie Brown
11.6k12938
11.6k12938
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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.
â Hurkyl
Jan 29 '14 at 5:23
Oh right. So is my definition correct, up to homotopy?
â 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. :(
â 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. :(
â 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.
â Ronnie Brown
Feb 2 '14 at 23:10