Let $K$ be the 1-simplex $\{v_0,v_1,[v_0,v_1]\}$.
I am trying to compute the reduced homology $\tilde{H}(K;\mathbb{R})$ from definition.
Q1) Is the following correct?
The general 0-cochain is of the form $c^0=r_0v_0^*+r_1v_1^*$. $$\langle\delta_0c^0,[v_0,v_1]\rangle=r_1-r_0.$$
Hence if $c^0$ lies in the kernel $\ker\delta_0$, then $r_0=r_1$. Hence, $ker\delta_0=\langle v_0^*+v_1^*\rangle\cong\mathbb{R}$.
Let $\epsilon^*(f)\in\text{Im}\,\epsilon^*$, where $f\in\text{Hom}(\mathbb{Z},\mathbb{R})$. Then $\langle\epsilon^*(f),[v_0]\rangle=\langle f,\epsilon[v_0]\rangle=\langle f,1\rangle$.
Similarly $\langle\epsilon^*(f),[v_1]\rangle=\langle f,\epsilon[v_1]\rangle=\langle f,1\rangle$.
I am not very sure how to proceed from here. Is it correct to say that $\text{Im}\,\epsilon^*\cong\langle v_0^*+v_1^*\rangle\cong\mathbb{R}$ also, since $\epsilon^*(f)$ has the form $\langle f,1\rangle(v_0^*+v_1^*)$?
If that is so, then I can see that $\tilde{H^0}(K;\mathbb{R})=0$.
Q2) In general, is there a theorem saying that for connected simplicial complex, the reduced cohomology $\tilde{H^0}$ is always zero?
Thanks.