Computing the Simplicial Homology of $S^1$ I have been hunting for a totally algebraic walkthrough of computing the simplicial homology of the circle, but I have only come across sources that simply triangulate $S^1$ and then geometrically observe the n-dimensional holes. Please provide me with a detailed computation of the "cycles modulo boundaries." 
 A: Let us consider $S^1$ as a triangle, with vertices $u$, $v$, and $w$ and edges between each pair of them which we will write as $uv$, $vw$, and $uw$.  We use the ordering $u<v<w$ on the vertices.  So we have $3$ $1$-simplices, $3$ $0$-simplices, and no simplices in any other dimension.  It follows that $H_n(S^1)=0$ for $n>1$, since there are no nonzero $n$-chains at all (so the cycle and boundary groups are both trivial).
To compute $H_1(S^1)$ and $H_0(S^1)$, we'll need to compute the boundary map $\partial_1:C_1\to C_0$.  The chain group $C_1$ has basis $\{uv,vw,uw\}$ and the chain group $C_0$ has basis $\{u,v,w\}$.  We have $\partial_1(uv)=v-u$, $\partial_1(vw)=w-v$, and $\partial_1(uw)=w-u$. We can thus compute the kernel of $\partial_1$: for $a,b,c\in\mathbb{Z}$, $$\partial_1(auv+bvw+cuw)=a(v-u)+b(w-v)+c(w-u)=-(a+c)u+(a-b)v+(b+c)w.$$ Thus $auv+bvw+cuw \in\ker(\partial_1)$ iff $a=b=-c$.  That is, $\ker(\partial_1)$ is isomorphic to $\mathbb{Z}$, generated by $uv+vw-uw$.  Since there are no $2$-simplices, the image of $\partial_2$ is $0$, so $H_1(S^1)\cong \mathbb{Z}/0\cong\mathbb{Z}$.
The last boundary map $\partial_0$ is $0$, so every $0$-chain is a cycle and $H_0(S^1)=C_0/\operatorname{im}(\partial_1)$.  The image of $\partial_1$ is generated by $v-u$, $w-v$, and $w-u$.  So $u$, $v$, and $w$ are all equal modulo the image of $\partial_1$, and so the quotient group is cyclic, generated by the coset of $u$.  On the other hand, no nonzero multiple of $u$ is in the image of $\partial_1$, since if $nu=\partial_1(auv+bvw+cuw)$, by the computation above we have $a+c=-n$, $a-b=0$ and $b+c=0$, which is impossible (the last two equations force $a+c=0$).  Therefore the coset of $u$ freely generates the quotient group $C_0/\operatorname{im}(\partial_1)$.  Thus $H_0(S^1)\cong\mathbb{Z}$.
