Computing the 0-dim cohomology group of a connected simplicial complex with coefficients in $\mathbb{Z}_{2}$

Question

Let $L$ be a connected locally finite simplicial complex with coefficients in $\mathbb{Z}_{2}$. I want to prove that $H^{0}(L)=\mathbb{Z}_{2}$. We know that $H^{0}(L)$= $Ker\delta_{1} / Im\delta_{0}=Ker\delta_{1}$ and $\delta_{1}: Hom(C_{0},\mathbb{Z}_{2})\to Hom(C_{0},\mathbb{Z}_{2})$ with $\delta_{1}(f)=fd_{1}$ ($d_{1}$ the boundary). I also know that we can see the elements of $Hom(C_{m},\mathbb{Z}_{2})$ i.e. the $m$-dim cochains, as the subsets of the set of all m-simplices but I am not sure how can I use that to prove the statement. Any help? Why $Ker\delta_{1}\cong \mathbb{Z}_{2}$? Also if $L$ has infinitely many $0$-simplices why $H_{f}^{0}=0$? ($H_{f}^{0}$ the restriction in finite cocycles modulo coboundaries of finite cochains.) Thank you in advance.

Answer 1

In the cochain complex $$ 0 \to \text{Hom}(C_0(L), \mathbb{Z}/2\mathbb{Z}) \to \text{Hom}(C_1(L), \mathbb{Z}/2\mathbb{Z}) \to \dots,$$ the group of 0-chains $C_0(L)$ is simply the free abelian group generated by all points of $L$, and so $\text{Hom}(C_0(L), \mathbb{Z}/2\mathbb{Z}) = \text{Fun}(X,\mathbb{Z}/2\mathbb{Z})$. For $\phi \in \text{Hom}(C_0(L), \mathbb{Z}/2\mathbb{Z})$ and $\sigma: \Delta^1 \to L$ a 1-simplex, we have $$ \delta^1 \phi (\sigma) = \phi (\partial \sigma) = \phi (\sigma(v_1)) - \phi(\sigma(v_0)),$$ so in other words $\phi \in \ker \delta^1$ if and only if $$ \phi(\gamma(1)) = \phi(\gamma(0))$$ for all paths $\gamma: [0,1] \to L$.

In this way, $H^0(L;\mathbb{Z}/2\mathbb{Z})$ is the space of all functions $L \to \mathbb{Z}/2\mathbb{Z}$ which are constant on every path component. In particular then, if $L$ is path connected then its zeroth cohomology group is $\mathbb{Z}/2\mathbb{Z}$.

Note that none of this required the simplicial complex structure.