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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: enter image description here

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!

Henry T.
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2 Answers2

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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.

Eric Wofsey
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Ignoring labels for the moment, the standard picture of $\Delta^2$ is always oriented in this way: $v_0\to v_1\to v_2$, in the manner of image $A$. So this is what Hatcher wanted you to do.

We can make this formal without seemingly unrigorously - it's fine, but maybe confusing for you - talking about orientations and arbitrary labels like $v_0$ and $e_0$ etc. in two different ways.

A $\Delta$-complex structure is really just a homeomorphism between that space and the geometric realisation of some simplicial $\Delta$-set. Here, Hatcher wants a $\Delta$-set $X$ with $X_2$ containing $(n+1)$ elements, say, $\tau_0,\tau_1,\cdots,\tau_n$. When we identify the edges, what we want to say is that: $d_0\tau_0=d_1\tau_0=d_2\tau_0$ and also to say that $d_0\tau_j=d_2\tau_j$ for $1\le j\le n$ and $d_1\tau_j=d_2\tau_{j-1}$. To make everything work we equip $X_0$ with one vertex only. It's easy to see we can construct a $\Delta$-set meeting these criteria, and then we care about the homology of $|X|$ (which may as well be computed as the simplicial homology of $X$).

Alternatively we can look at the space $Y$ which is just $\bigsqcup_{j=0}^{n+1}|\Delta^2|$ and take the quotient by the following relation: (using barycentric coordinates) $(t_1,t_2,0;0)\sim(t_1,0,t_2;0)\sim(0,t_1,t_2;0)$ and $(t_1,t_2,0;j)\sim(0,t_1,t_2;j)$ and $(t_1,0,t_2;j)\sim(t_1,t_2,0;j-1)$. Then you have to check the maps $|\Delta^2|\hookrightarrow Y\twoheadrightarrow X$, where $X$ is the quotient space, define a $\Delta$-structure on $X$. In this case they do and you can compute simplicial homology.

Both formalisms could even be phrased as a colimit.

FShrike
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