0
$\begingroup$

Given a simplicial set $X$, we define $\pi_0(X)$ to be the coequalizer of $d_0, d_1: X_1 \to X_0$. I have to show that $\pi_0(X) \cong \pi_0( | X| )$. Does this mean that we are taking $X_0/ \sim$, where $x \sim y$ if the two vertices can be connected by a chain of $1-$simplices?

I thought that maybe a good idea would be to write $X$ as a disjoint union of its path-connected components (somehow?), so that we can restrict to this case of $X$ path-connected, getting both $\pi_0(X)$ and $\pi_0(|X|)$ to have only a single element. Or maybe it makes more sense to prove explicitly that there is a bijection?

Finally, given an abelian group $A$, I would like to show that there is an isomorphism $H_0(X, A) \cong \oplus_{\pi_0(X)}A$. I'm actually a little bit confused by homology with coefficients. It makes sense that, if we decompose $X$ into its path-connected components $X_i$, then we get an isomorhism $H_0(X) \cong \oplus_i H_0(X_i)$. Does the result with coefficients simply come from this?

Thank you for your help.

$\endgroup$
5
  • $\begingroup$ check the kerodon to see why simplicial connected components and topological connected components are the same. really it's a result about CW complexes and when graphs (1-dimensional complexes) are path-connected $\endgroup$
    – FShrike
    Dec 6, 2023 at 17:50
  • $\begingroup$ @FShrike Thank you, I will have a look at the Kerodon for the first part then. Do you have any advice for the second question? $\endgroup$
    – idontknow
    Dec 6, 2023 at 21:58
  • $\begingroup$ What kind of homology are you using? For any chain complex $C_\ast$ concentrated in nonnegative degrees we have $H_0(C_\ast\otimes A) = H_0(C_\ast)\otimes A$, so for example for singular homology you have $H_0(X;A)\cong H_0(X)\otimes A$. $\endgroup$ Dec 6, 2023 at 22:28
  • $\begingroup$ @VincentBoelens I'm working with simplicial homology. $\endgroup$
    – idontknow
    Dec 6, 2023 at 22:31
  • $\begingroup$ @idontknow Does my answer for the second part make sense? $\endgroup$
    – FShrike
    Dec 12, 2023 at 19:41

1 Answer 1

0
$\begingroup$

There is a more simplicial/category - theoretic way of understanding the connected components thing. It's in my notes but it was a while ago, so just read the Kerodon. A supplementary resource for you is my old answer here. This is a (long-winded and conceptual) explanation of $\bigoplus_{\pi_0(X)}A\cong H_0(X; A)$ for simplicial sets $X$. It doesn't say that but what you want follows very quickly; the key thing is that the zeroth homology is generated over $A$ by all vertices $v$ modulo $v\sim w$ if $v=d_0e,\,w=d_1e$ for some edge $e$ or visa versa, and that coequaliser $X_1\overset{d_0,d_1}{\rightrightarrows}X_0$ is isomorphic to $\pi_0(X)$. The Kerodon explains this (in less detail?) and also covers $\pi_0(X)\cong\pi_0(|X|)$.

Here's a topological way to understand $\pi_0(X)\cong\pi_0(|X|)$ (which may turn out to be a very similar argument to the one presented in the Kerodon).

It is known for CW complexes $Y$ that $\pi_0(Y)\cong\pi_0(Y_1)$ where $Y_1$ is the $1$-skeleton, which is nothing but a graph which is nothing but (! the Kerodon has a nice subchaper on this) (the realisation of) a $\le1$-dimensional simplicial set $X$. Well, if $Y=|X|$ then $Y_1$ is nothing but $|X_1|$, so it suffices to think about $X$'s $1$-skeleton. To justify our claim, we need to say that topological connectivity of a graph is really the same notion as edgewise, "combinatorial" connectivity of a graph (which is, by the $X_1\overset{d_0,d_1}{\rightrightarrows}X_0\twoheadrightarrow\pi_0(X)$ coequaliser theorem, nothing but the connectivity notion for $X$ or indeed $X_1$ as a simplicial set).

If the topological graph $G$ is connected in the combinatorial sense, it is easy to see it is topologically connected since the edges themselves offer you a path.

Conversely, say the topological graph $G$ is connected in the topological sense. Let $v,v'$ be any two vertices. Take a continuous path between them. Its image is compact hence contained in a finite subcomplex hence contained in finitely many edges; remove redundancies i.e. edges that are disjoint from the image. An edge $e_0$ contains $v$. Say $v'$ is not a vertex of $v$. If no other edges from this finite set meet $e_0$, we have a contradiction since the finite union of closed things is closed and it follows $e_0$ is clopen in the subgraph of these finitely many edges, and as such the path must be wholly contained in $e_0$. So, some edge $e_1$ meets $e_0$. If $v'$ is not a vertex of $e_1\cup e_0$, we rinse and repeat until we have exhausted all the finitely many edges. $v'$ is a vertex of at least one of these, so we will eventually find an edgewise connection.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .