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Stream: learning: questions

Topic: degree 0 elements of the nerve

Tim Hosgood (Aug 02 2021 at 19:44):

When we look at the nerve of a ($1$-)category, the $n$-simplices are chains of composable morphisms of length $n$... except for $n=0$, when we simply have objects. Is there any "harm" in modifying the definition so that the $0$-simplices are instead the identity morphisms of the category? I can't tell if this actually is different from the usual definition or whether it's entirely identical... but I've never seen it before, so I'm assuming that it's not very nice in some way

Nathanael Arkor (Aug 02 2021 at 19:52):

This is exactly the same, because objects are in bijection with identity morphisms.

Tim Hosgood (Aug 02 2021 at 19:54):

that's what I was thinking, so was just wondering why it's not the usual definition, since then every simplex is a chain of morphisms, rather than having this little caveat for $n=0$

Nathanael Arkor (Aug 02 2021 at 19:56):

I don't know the reason, but I would guess it's for the same reason that most people don't give the unsorted morphisms-only definition of a category – though in this case, I agree it seems conceptually simpler to avoid this caveat.

Reid Barton (Aug 02 2021 at 20:08):

A chain of $n$ composable morphisms involves $n+1$ objects, so it doesn't seem exceptional that for $n=0$ you get a single object (and no morphisms).

Fawzi Hreiki (Aug 02 2021 at 20:24):

Exactly, an object is just a path of length 0.

Ian Coley (Aug 02 2021 at 20:28):

Going back to the idea of a simplicial set as the building blocks of a topological space/simplicial complex, the 0-simplices are vertices, which we don't want to think of as having any edges. That is thinking very rigidly, but sometimes that's nice!

Tim Hosgood (Aug 02 2021 at 20:34):

Reid Barton said:

A chain of $n$ composable morphisms involves $n+1$ objects, so it doesn't seem exceptional that for $n=0$ you get a single object (and no morphisms).

this is true, and is a very good reason

Patrick Nicodemus (Aug 04 2021 at 20:52):

Tim Hosgood said:

that's what I was thinking, so was just wondering why it's not the usual definition, since then every simplex is a chain of morphisms, rather than having this little caveat for $n=0$

I think of $\Delta$ as having an inclusion functor $i$ into $\mathbf{Cat}$, because posets are a special type of categories. Then the nerve is just given by $X\mapsto Hom(i(-), X)$. Composable strings of morphisms are an abbreviated notation for functors from these poset categories.

ADITTYA CHAUDHURI (Aug 04 2021 at 21:38):

Tim Hosgood said:

When we look at the nerve of a ($1$-)category, the $n$-simplices are chains of composable morphisms of length $n$... except for $n=0$, when we simply have objects. Is there any "harm" in modifying the definition so that the $0$-simplices are instead the identity morphisms of the category? I can't tell if this actually is different from the usual definition or whether it's entirely identical... but I've never seen it before, so I'm assuming that it's not very nice in some way

So you are saying that if we identify the elements of $N_0$ with the image of the degeneracy map $s^{0}:N_0 \rightarrow N_1$ then are we loosing any information or not. But, using degeneracy maps , I guess in that sense we can also identify the elements of $N_1$ with the elements of $N_2$ and so on. I am not sure about what will happen if we do this.

ADITTYA CHAUDHURI (Aug 04 2021 at 21:57):

Basically in that sense , any morphism $f$ in a usual category $C$ has to be written as $f \circ id \circ id \circ......\circ id \circ......$. My point is that if we want to do it for $N_0$ then why not for $N_1$, and one may ask this question repeatedly for higher indices. Am I misunderstanding anything here?

ADITTYA CHAUDHURI (Aug 04 2021 at 22:04):

I guess the left adjoint of the nerve functor, the fundamental category functor $\tau_1: \rm{sSets} \rightarrow \rm{Cat}$ exactly identifies "these arrows" to form an ususal category .