# Hatcher 2.1.6

Hatcher, Algebraic Topology, Chapter 2, Section 1

6. Compute the simplicial homology groups of the $\Delta$-complex obtained from $n+1$ 2-simplices $\Delta^2_0,\ldots,\Delta^2_n$ by identifying all three edges of $\Delta^2_0$ to a single edge, and for $i>0$ identifying the edges $[v_0,v_1]$ and $[v_1,v_2]$ of $\Delta^2_i$ to a single edge and the edge $[v_0,v_2]$ to the edge $[v_0,v_1]$ of $\Delta^2_{i-1}$.
Proof.

$\square$

## One thought on “Hatcher 2.1.6”

1. Dom

My solution:

Let a_i, b_i, c_i be the edges [v_0, v_1], [v_1, v_2], [v_0, v_2] respectively of each \Delta_i^2. Then a_0 = b_0 = c_0, a_i = b_i, and c_i = a_{i-1} for all i. One can show inductively that all vertices are identified to a single vertex v. If \del is the boundary homomorphism, then \del_0 = \del_1 = 0. Since there’s only one vertex, it follows that H_0 = Z.

Also, by the identifications, the set of 1-boundaries is generated by [a_0, a_1, … , a_n]. One can compute that if \Delta_i^2 is the ith simplex, then \del_2(\Delta_0^2) = a_0, and \del_2(\Delta_i^2) = 2a_i – a_{i-1} for all i. Thus, the image is generated by [a_0, 2a_1 – a_0, … , 2a_n – a_{n-1}] = [a_0, 2a_1, 2a_2 – a_1, … , 2a_n – a_{n-1}].

Since \del_1 = 0, the kernel is [a_0, … , a_n]. So this representation of the image implies 2a_2 = a_1 in the quotient, and also 2a_1 = 0. Thus 4a_2 = 0. Inductively, 2^k a_i = 0. Conclusion: H_1 = Z/(2^n)Z.

With the generators a_i, one can show that the kernel of \del_2 is 0, so H_2 = 0. Of course, H_k = 0 for k > 2.

Thoughts?