# Connected components of a simplicial set and homology

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?

• 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 Dec 6, 2023 at 17:50
• @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? Dec 6, 2023 at 21:58
• 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 $H_0(X;A)\cong H_0(X)\otimes A$. Dec 6, 2023 at 22:28
• @VincentBoelens I'm working with simplicial homology. Dec 6, 2023 at 22:31
• @idontknow Does my answer for the second part make sense? Dec 12, 2023 at 19:41

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.