Presentation of $hS$ as in page 29 of HTT
up vote
1
down vote
favorite
At the page claims 29 of HTT, Lurie claims that the category $hS$ (we ignore all the enrichement) admits the following presentation by generators and relation :
The objects of $hS$ are the vertices of $S$ (this one is clear).
For every edge $ f: Delta^1 rightarrow S$ there is a morphism $overline{f}$ in $hS$.
For each $2$-simplex $sigma$ we have $overline{d_0sigma} circ overline{d_2sigma} = overline{d_1sigma}$.
For each vertex $x$ the morphism $overline{s_0x}$ is the identity $id_x$.
He claims that this follows from the fact the functor $h$ is left adjoint to the nerve functor but I don't see how.
We first idea was to use Yoneda and try to use the adjunction on $sSet( Delta^1, S)$ to get the statement about morphisms but the adjunction isnot in the correct direction.
Then I tought about using that $hDelta^1 cong [1]$ as a category and use that that the set of morphisms in $hS$ is in bijection with $Cat([1], hS)$ but I did not get much result.
combinatorics category-theory higher-category-theory
asked Nov 15 at 15:21
C.Montes
343
add a comment |
up vote
1
down vote
favorite
At the page claims 29 of HTT, Lurie claims that the category $hS$ (we ignore all the enrichement) admits the following presentation by generators and relation :
The objects of $hS$ are the vertices of $S$ (this one is clear).
For every edge $ f: Delta^1 rightarrow S$ there is a morphism $overline{f}$ in $hS$.
For each $2$-simplex $sigma$ we have $overline{d_0sigma} circ overline{d_2sigma} = overline{d_1sigma}$.
For each vertex $x$ the morphism $overline{s_0x}$ is the identity $id_x$.
He claims that this follows from the fact the functor $h$ is left adjoint to the nerve functor but I don't see how.
We first idea was to use Yoneda and try to use the adjunction on $sSet( Delta^1, S)$ to get the statement about morphisms but the adjunction isnot in the correct direction.
Then I tought about using that $hDelta^1 cong [1]$ as a category and use that that the set of morphisms in $hS$ is in bijection with $Cat([1], hS)$ but I did not get much result.
combinatorics category-theory higher-category-theory
asked Nov 15 at 15:21
C.Montes
343
It would be helpful if you could edit your $LaTeX$. Anyway, the point is that $h$ is cocontinuous. So this is immediately true for the 2-skeleton of $S$ via its decomposition as a colimit of simplices. And the higher-dimensional simplices don't affect anything here. More formally, you can check that $h$ sends the $n$-sphere and the $n+1$-simplex to the same category for $ngeq 2$.
– Kevin Carlson
Nov 15 at 17:54
It's funny because nothing was wrong, it came out perfectly under the box where I wrote my code. I just resubmitted it and now it's fine.
– C.Montes
Nov 15 at 19:00
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
At the page claims 29 of HTT, Lurie claims that the category $hS$ (we ignore all the enrichement) admits the following presentation by generators and relation :
The objects of $hS$ are the vertices of $S$ (this one is clear).
For every edge $ f: Delta^1 rightarrow S$ there is a morphism $overline{f}$ in $hS$.
For each $2$-simplex $sigma$ we have $overline{d_0sigma} circ overline{d_2sigma} = overline{d_1sigma}$.
For each vertex $x$ the morphism $overline{s_0x}$ is the identity $id_x$.
He claims that this follows from the fact the functor $h$ is left adjoint to the nerve functor but I don't see how.
We first idea was to use Yoneda and try to use the adjunction on $sSet( Delta^1, S)$ to get the statement about morphisms but the adjunction isnot in the correct direction.
Then I tought about using that $hDelta^1 cong [1]$ as a category and use that that the set of morphisms in $hS$ is in bijection with $Cat([1], hS)$ but I did not get much result.
combinatorics category-theory higher-category-theory
asked Nov 15 at 15:21
C.Montes
343
At the page claims 29 of HTT, Lurie claims that the category $hS$ (we ignore all the enrichement) admits the following presentation by generators and relation :
The objects of $hS$ are the vertices of $S$ (this one is clear).
For every edge $ f: Delta^1 rightarrow S$ there is a morphism $overline{f}$ in $hS$.
For each $2$-simplex $sigma$ we have $overline{d_0sigma} circ overline{d_2sigma} = overline{d_1sigma}$.
For each vertex $x$ the morphism $overline{s_0x}$ is the identity $id_x$.
He claims that this follows from the fact the functor $h$ is left adjoint to the nerve functor but I don't see how.
We first idea was to use Yoneda and try to use the adjunction on $sSet( Delta^1, S)$ to get the statement about morphisms but the adjunction isnot in the correct direction.
Then I tought about using that $hDelta^1 cong [1]$ as a category and use that that the set of morphisms in $hS$ is in bijection with $Cat([1], hS)$ but I did not get much result.
combinatorics category-theory higher-category-theory
combinatorics category-theory higher-category-theory
asked Nov 15 at 15:21
C.Montes
343
asked Nov 15 at 15:21
C.Montes
343
edited Nov 15 at 18:59
asked Nov 15 at 15:21
C.Montes
343
asked Nov 15 at 15:21
C.Montes
343
asked Nov 15 at 15:21
C.Montes
343
343
It would be helpful if you could edit your $LaTeX$. Anyway, the point is that $h$ is cocontinuous. So this is immediately true for the 2-skeleton of $S$ via its decomposition as a colimit of simplices. And the higher-dimensional simplices don't affect anything here. More formally, you can check that $h$ sends the $n$-sphere and the $n+1$-simplex to the same category for $ngeq 2$.
– Kevin Carlson
Nov 15 at 17:54
It's funny because nothing was wrong, it came out perfectly under the box where I wrote my code. I just resubmitted it and now it's fine.
– C.Montes
Nov 15 at 19:00
add a comment |
It would be helpful if you could edit your $LaTeX$. Anyway, the point is that $h$ is cocontinuous. So this is immediately true for the 2-skeleton of $S$ via its decomposition as a colimit of simplices. And the higher-dimensional simplices don't affect anything here. More formally, you can check that $h$ sends the $n$-sphere and the $n+1$-simplex to the same category for $ngeq 2$.
– Kevin Carlson
Nov 15 at 17:54
It's funny because nothing was wrong, it came out perfectly under the box where I wrote my code. I just resubmitted it and now it's fine.
– C.Montes
Nov 15 at 19:00
It would be helpful if you could edit your $LaTeX$. Anyway, the point is that $h$ is cocontinuous. So this is immediately true for the 2-skeleton of $S$ via its decomposition as a colimit of simplices. And the higher-dimensional simplices don't affect anything here. More formally, you can check that $h$ sends the $n$-sphere and the $n+1$-simplex to the same category for $ngeq 2$.
– Kevin Carlson
Nov 15 at 17:54
It would be helpful if you could edit your $LaTeX$. Anyway, the point is that $h$ is cocontinuous. So this is immediately true for the 2-skeleton of $S$ via its decomposition as a colimit of simplices. And the higher-dimensional simplices don't affect anything here. More formally, you can check that $h$ sends the $n$-sphere and the $n+1$-simplex to the same category for $ngeq 2$.
– Kevin Carlson
Nov 15 at 17:54
It's funny because nothing was wrong, it came out perfectly under the box where I wrote my code. I just resubmitted it and now it's fine.
– C.Montes
Nov 15 at 19:00
It's funny because nothing was wrong, it came out perfectly under the box where I wrote my code. I just resubmitted it and now it's fine.
– C.Montes
Nov 15 at 19:00
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999820%2fpresentation-of-hs-as-in-page-29-of-htt%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
It would be helpful if you could edit your $LaTeX$. Anyway, the point is that $h$ is cocontinuous. So this is immediately true for the 2-skeleton of $S$ via its decomposition as a colimit of simplices. And the higher-dimensional simplices don't affect anything here. More formally, you can check that $h$ sends the $n$-sphere and the $n+1$-simplex to the same category for $ngeq 2$.
– Kevin Carlson
Nov 15 at 17:54
It's funny because nothing was wrong, it came out perfectly under the box where I wrote my code. I just resubmitted it and now it's fine.
– C.Montes
Nov 15 at 19:00