# The zeroth homology group corresponds to path components

I'm doing an independent study on homology theory and have noticed that the rank of $H_0(X)$ is the number of path components in $X$. I'm trying to prove this with simplicial homology, but I keep getting stuck because I'm not sure how structure my proof and thus do not know where to start.

Hatcher gives a proof in his book, but it is very dense and I'm having a hard time following the logic (and his notation since I haven't been studying from Hatcher as he is too dense for my current level).

I'm hoping someone can help me develop a nice formal proof that is understandable for an undergraduate who has taken both topology and abstract algebra.

The idea behind is the following: suppose $X$ is path connected. $H_0(X)$ is the quotient of the zero cycles $Z_0(X)=\{n_0x_0+...+n_xx_p, x_0,...,x_p\in X\}$ by the image $d_1:C_1(Z)\rightarrow Z_0(X)$ where $C_1(X)$ is module of $1$-chains.

Let $x,y\in X$, there exists a path $c:I=[0,1]\rightarrow X$ such that $c(0)=x,c(1)=y$. $c(I)$ is a $1$-chain and $d_1((c(I))=x_1-x_0$, this implies that $Z_0(X)/d_1(C_1(X))$ is generated by the class of $x$.

• You have shown one direction: that $X$ is path connected implies $H_0(X)\cong\mathbb{Z}$. Here is my attempt at the other direction. Suppose $H_0(X)\cong\mathbb{Z}$. Then there is only one homology class for the vertices of $X$. So for any two vertices $v,w$, there is $\sigma\in C_1(X)$ where $v-w=\partial_1\sigma$. Thus $\sigma$ is a path between $v$ and $w$. Therefore $X$ is path connected. Is it okay to say that $v-w=\partial_1\sigma$ implies $\sigma$ is a path from $v$ to $w$? – rosterherik May 9 '18 at 17:22
• Yes, you can always realize a $1$-simplex by a path. – Tsemo Aristide May 9 '18 at 17:25

Thanks to nudges by Tsemo Aristide, I was able to write a proof I am happy with, so I will post it here for others who may be curious.

We will first prove that $K$ is path connected if and only if $H_0(K)\cong\mathbb{Z}$.

$(\Rightarrow)$ Suppose that $K$ is path connected. Fix a vertex $v$ in $K$. Then for any other vertex $w$ in $K$, there is a path from $v$ to $w$ in $K$. Since $K$ is a simplicial complex, this path is necessarily a sequence of edges in $K$ of the form $\sigma=(v,v_1)+(v_1,v_2)+\cdots+(v_k,w)$. Then $\partial_1 \sigma=w-v$. Thus $v$ and $w$ are homologous vertices. Since this holds for any vertex $w$ of $K$, we have that all vertices in $K$ are homologous and so there is only one homology class for the vertices of $K$. Therefore $H_0(K)=\langle v\rangle\cong\mathbb{Z}$.

$(\Leftarrow)$ Suppose that $H_0(K)\cong\mathbb{Z}$. Then there is only one homology class on the vertices of $K$. So, for any pair of vertices $v,w$ of $K$, there is some $\sigma\in C_1(K)$ where $\partial_1\sigma=w-v$. Without loss, we may assume that $\sigma$ is an elementary $1$-chain. Then $\sigma$ is a path from $v$ to $w$, and so $K$ is path connected.

Suppose now that $K$ has path components $P_i$, where $1\leq i\leq\ell$. It follows from $(\Rightarrow)$ that $H_0(P_i)\cong\mathbb{Z}$ for each $i$. It follows from $(\Leftarrow)$ that $H_0(P_i\cup P_j)\cong\mathbb{Z}^2$ for any $i,j$ where $i\neq j$. Therefore, since $K=\bigcup_{i=1}^{\ell} P_i$, we have $H_0(K)\cong\mathbb{Z}^{\ell}$.