Let $K$ be an (abstract) simplicial complex. The claim is: $H_0(K;\mathbb{Z})$ is always nonzero. Is this possible to prove it without any "special techniques to computing homology-groups"? $H_0(K;\mathbb{Z})=\mathrm{ker}(\delta_0)/\mathrm{im}(\delta_1)$, $\delta_n$ is the boundary operator and $\delta_0:=0$. Therefore I have to compute $C_0(K;\mathbb{Z})=\mathbb{Z}[\Sigma_0]=\mathrm{ker}(\delta_0)$, $\Sigma_0$ is the set of the 0-simplices. And $\delta_1\colon C_1(K;\mathbb{Z})\to C_0(K;\mathbb{Z})$. I only know how to compute homology groups of simple examples of a simplicial complex. Maybe I have to wait if I get more information in lecture next week. Or is it possible to prove the claim without any useful theorems and only with the definition of homology group? Regards

  • 1
    $\begingroup$ Well, it's not always nontrivial. If your notion of simplicial complex allows for the empty complex, then the 0th homology can be trivial. But provided your complex is non-empty yes, you can prove it is always non-trivial. Look up the definition of reduced homology and the proof that the reduced homology has rank one less than the normal homology (in dimension zero). $\endgroup$ Oct 23, 2014 at 22:52
  • $\begingroup$ Note that the boundary of a $1$-chain always has even parity. [The boundary of a $1$-simplex is of the form $1\cdot e + (-1)\cdot s$.] Thus any $0$-cahin of odd parity is not a boundary. Since, if the complex isn't empty, there are always $0$-chains of odd parity ($1\cdot p$), the quotient $H_0(K;\mathbb{Z})$ is nontrivial. $\endgroup$ Oct 23, 2014 at 23:13
  • $\begingroup$ thank you. Ryan, i dont know the reduced homology, but thank you for your idea. i will try to understand it. and Daniel: thank you. i understand it but i needed a few days^^. $\endgroup$
    – user151465
    Oct 29, 2014 at 9:06

1 Answer 1


The short answer is that $\mathrm{im}\partial_1$ is contained in the subgroup $B$ of $C_0$ given by

$\{\sum \lambda_i v_i : \sum \lambda_i = 0, \lambda_i \in \mathbb{Z}, v_i \in K_0\}$

This subgroup is proper as soon as $K_0$ is non-empty: $v \notin B$ for $v \in K_0$.

The long story is that you can slightly change your simplicial complex by adding a single $-1$ simplex, $*$. So $C_{-1} = \mathbb{Z}[*]$, and define the boundary of any vertex simplex to be $*$.

This makes more sense than it seems to at first. The definition of a simplicial complex says it's a "family of non-empty finite sets closed under the operation of taking non-empty subsets" [wikipedia]. If you get rid of the totally unnecessary requirement that the subsets be non-empty, you get the $-1$ simplex corresponding to the empty set naturally.

Now the boundary maps of satisfy the face identity: $d_i \circ d_j = d_{j-1} \circ d_i$ for $i < j$.

In other words taking a simplex $[0,1...,n]$ and omitting the $j$th vertex and then the $i$th gives you $[0,1,..\hat{i}...\hat{j}...n]$. This is the same as omitting the $i$th vertex first, and then in the new simplex $[0,1,...\hat{i}...n]$ omitting the $j-1$st vertex.

The face identity is the key ingredient in showing that $\partial^2 = 0$! So the same argument tells you that after adding $C_{-1}$ to you complex is still going to have $\partial^2 = 0$.

If your simplicial complex is non-empty it has at least one vertex and that doesn't get mapped to 0 any more. It's sent to $*$. This says that the zeroth homology of the extended complex $\tilde{H}_0$ is strictly smaller than the original homology $H_0$.

  • $\begingroup$ thank you, i understand it. i see, there are lots of possibilities to prove this. i have an other proof now, but alternative ways are always good. $\endgroup$
    – user151465
    Oct 29, 2014 at 9:11

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