I have a question regarding exercise 2.1.6 in Hatcher:
Compute the simplicial homology groups of the $\Delta$-complex obtained from $n+1$ $2$-simplices $\Delta_0^2,…,\Delta_n^2$ by identifying all three edges of $\Delta_0^2$ to a single edge, and for $i>0$ identifying the edges $[v_0,v_1]$ and $[v_1,v_2]$ of $\Delta_i^2$ to a single edge and the edge $[v_0, v_2]$ to the edge $[v_0, v_1]$ of $\Delta_{i-1}^2$.
The answer has already been discussed f.e. here: Hatcher exercise 2.1.6 (Simplicial homology)
In the answer by FShrike in the question linked above it is stated that: $\partial(\Delta_0^2)\sim e_0-e_0+e_0=e_0$. I get this if the edges are identified as in A in my sketch below:
Because then $\partial(\Delta_0^2) = \partial [v_0,v_1,v_2]=[v_0,v_1]-[v_0,v_2]+[v_0,v_1]=e_0-e_0+e_0$.
But if they were identified as in B one would get: $\partial(\Delta_0^2) = \partial [v_0,v_1,v_2]=[v_0,v_1]-[v_0,v_2]+[v_0,v_1]=e_0+e_0+e_0=3e_0$. Which, down the line, would lead to a different homologous group. My question is, how do I know how to identify these edges, i.e. in „what direction to glue them together“?
Thank you!
Best Answer
Each face of a simplex in a $\Delta$-complex is required to be another simplex of the $\Delta$-complex, in a way that preserves the ordering of the vertices. In particular, then, if you have a $2$-simplex with vertices $v_0,v_1,v_2$ (in that order), its three boundary edges must be $1$-simpleces of the $\Delta$-complex structure, where the vertices of each boundary edge are ordered in the order $v_0,v_1,v_2$ (except with one of the vertices omitted).
In the example in question, then, the one edge $e_0$ must be equal to all three boundary edges of the triangle, with the vertex ordering $v_0,v_1,v_2$. That means the picture looks like $A$ rather than $B$. Indeed, $B$ is not even a valid $\Delta$-complex at all: it would be trying to have two different edges that are identified as the reverse of each other, which is not allowed.