Connected Component of a simplicial set

Question

From this note, I'm trying to solve 6.8: https://faculty.math.illinois.edu/~rezk/quasicats.pdf

I will list the necessary definitions:

And $e_0$ refers to $X(f)(e)$, where $f \colon [0] \rightarrow [1]$ such that $f(0) = 0$, and $e_1$ is similarly $X(g)(e)$, where $g \colon [0] \rightarrow [1]$ such that $g(0) = 1$.

My attempt:

I am thinking the canonical bijection is sending $[a]_{\approx_1}$ to $[a]_{\approx}$.

To show that mapping is well-defined, it suffices to show $a \sim_1 b$ implies $a \approx b$.

But this is clear since as $e_0 = a$, $e_1 = b$, so $a \approx e$ and $b \approx e$, so $a \approx b$.

To show that mapping is surjective, let $a \in X_n$,and consider $<0> \colon [0] \rightarrow [n]$ defined by $<0>(0) = 0$.

Let's write $a_0$ for $X(<0>)(a)$.

Then $a \approx a_0$, so $[a_0]_{\approx_1}$ is mapped to $[a_0]_{\approx} = [a]_{\approx}$.

The part I'm having trouble with is showing that the mapping is injective. To show injectivity, we need that for $a, b \in X_0$ such that $a \approx b$, we need $a \approx_1 b$. But I have no idea how to show this.

Thanks for your help!

Answer 1

If $a,b\in X_0,\ a\approx b$, then there's a sequence $a=x_0,x_1,\dots,x_k=b$ of simplices ($x_i\in \bigsqcup_n X_n$) such that $x_i\sim x_{i+1}$ or $x_{i+1}\sim x_i$ for each $i=0,1,\dots,k-1$.

In both cases, $x_i$ and $x_{i+1}$ share a common vertex, say $y_i$, and inside one simplex $x_i$ any two vertex can be connected by edges (so all vertices of $x_i$ are in the same $\approx_1$ class), so there's a path of edges $a\leadsto y_0\leadsto y_1\dots\leadsto b$.

Answer 2

This is a slightly ramble-y exposition of the same idea with a different initial definition, for my own later benefit (so I can take paper notes when I have the time) and for anyone else's benefit who's a little confused (or who's starting with the definitions I start with).

Reference

The connected component can be defined rather differently. Fix a nonempty simplicial set $X$. A subobject $S\subseteq X$ is said to be a summand of $X$ when there exists a simplicial set $S^c$ that $S\sqcup S^c\cong X$ in the category of simplicial sets. A simplicial set $C$ is said to be connected when it is nonempty and it only admits the empty simplicial set or $C$ itself as summands, by analogy with topological connectivity (as phrased in terms of clopen subspaces). A connected component of $X$ is a summand $S$ of $X$ that is connected.

This is a reasonable definition, as the connected components of $X$ form a disjoint partition of $X$ and, if $\varnothing\neq S\subseteq C\subseteq X$ for $C$ a connected component of $X$, any connected subobject $T\subseteq X$ that contains $S$ shall have $T\subseteq C$, so $C$ is maximal. Proof: define $A=T\cap C$ and $B=T\cap C^c$, both subobjects of $X$. We are given that $S\subseteq A$ and $A,B$ are disjoint simplicial subsets of $X$ (in the sense that their categorical intersection is the empty simplicial set). Since $T$ is connected, either $A$ or $B$ are empty - but $S\subseteq A$ is nonempty, so $B$ is empty and $T=A\subseteq C$.

For a natural map $\varphi:U\implies V$ of simplicial sets, it can be shown that if $U$ is connected then the simplicial set $\varphi(U)\subseteq V$ is connected, and contained in a unique component of $V$. With this, for any $n\in\Bbb N_0$ and $\sigma\in V_n$ there is a natural transformation $\overline{\sigma}:\Delta^n\implies V$ - and $\Delta^n$ is connected, considering $1_n\in\Delta^n_n$ - so the image $\overline{\sigma}(\Delta^n)\subseteq V$ is connected and contained in the connected component of $\sigma$, $C_\sigma$.

Let $\tau\in V_m,\,\sigma\in V_n$. If there is $f\in\Delta(m,n)$ with $V_f(\sigma)=\tau$ - or, as OP writes, $\tau=\sigma f$ - then that is precisely the statement that $\tau=\overline{\sigma}(f)$, so $\tau\in\overline{\sigma}(\Delta^n)\subseteq C_{\sigma}$ and $\sigma\in C_{\tau}$. The equivalence relation generated by $\sigma\sim\tau$ (in the OP's notation) then carves out a subset of a connected component. By that I mean every equivalence class $[\sigma]$ is fully contained in $C_\sigma$, but to marry the OP's notions with the definitions above we need to show $[\sigma]=C_\sigma$. It is equivalent to show that $[\sigma]$ is a simplicial subset of $V$ and also a connected summand.

To show $[\sigma]$ is a simplicial subset of $V$, for $t:p\to r$ we need to show $V_t(\tau)\sim\sigma$ if $\tau\in X_r$ has $\tau\sim\sigma$. It suffices to show $\tau\sim V_t(\tau)$, but this is trivially true.

Suppose $S,T$ are simplicial sets with $S\sqcup T\cong[\sigma]$. As $\overline{\tau}(\Delta^m)$ is connected for any $\tau\in V_m$, each of these must fall as a subobject of either $S$ or $T$. That is, if $V_t(\tau)=\tau'$ or $V_t(\tau')=\tau$, $\tau,\tau'$ are forced to be both in either $S$ or $T$. But now that shows $[\sigma]$ falls entirely within either $S$ or $T$, so $[\sigma]$ is connected.

To show it is a summand, it suffices to show that $m\mapsto V_m\setminus[\sigma]_m$ is functorial, that is, if $\omega\in V_m\setminus[\sigma]_m$ and $f:k\to m$, then $V_f(\omega)\in V_k\setminus[\sigma]_k$. But that is also true since $V_f(\omega)\sim\sigma$ implies $\omega\sim V_f(\omega)\sim\sigma$, a contradiction.

I want to elaborate on Berci's assertion that:

"inside one simplex $x_i$ any two vertex can be connected by edges"

Since it confused me at first. Their argument can be recovered if we show that if $\sigma,\tau$ are vertices in the sub-simplicial set $\overline{x_i}(\Delta^{k_i})$ then they are edge-connected. We need to find a $1$-simplex $\epsilon=V_t(x_i)$ for some $t:1\to k_1$ with $d_1(\epsilon)=\sigma,d_0(\epsilon)=\tau$ or possibly the other way round. $\sigma=V_f(x_i),\,\tau=V_g(x_i)$ for some $f,g:0\to k_i$. We can identify these as simply integer points $0\le f,g\le k_i$. The arrow $t:=f\sqcup g:1\to k_i$ (if $f\le g$, or the other way round, if $g\le f$) satisfies $d_0(V_t(x_i))=V_{t\delta_0}(x_i)=V_g(x_i)=\tau$ and $d_1(V_t(x_i))=\sigma$, so indeed $\epsilon:=V_t(x_i)$ connects $\sigma\approx_1\tau$ "inside the simplex $x_i$".