# Hatcher 2.1.4

Hatcher, Algebraic Topology, Chapter 2, Section 1

4. Compute the simplicial homology groups of the triangular parachute obtained from $\Delta^2$ by identifying its three vertices to a single point.

Proof. The space $\Delta^2$, as well as the space $X$ obtained by the identification, can be seen in this (crude) sketch.

For the homology calculations, note that $\Delta^2$ has three vertices $v_i$, $i=1,2,3$, three edges, and one face (call it $F$). Under the given identification, the space $X$ will then have one vertex $\widehat{v}$, three edges $a,b,c$ (all of the form $[\widehat{v},\widehat{v}]$), and one face $F$. In particular, then, $C_0(X),C_2(X)\cong\mathbb{Z}$ and $C_1(X)=\langle a,b,a+b-c\rangle\cong\mathbb{Z}^3$. Moreover, we have the following chain complex1:

To determine the simplicial homology of $X$, it suffices to determine the actions of $\partial_i$, $i=1,2$. Note that $\partial_2$ maps the face $F$ to the linear combination $a+b-c$ of the edges and that $\partial_1$ maps any edge $a,b,c=[\widehat{v},\widehat{v}]$ to the linear combination $\widehat{v}-\widehat{v}=0$. Hence, $\text{Im}(\partial_2)=a+b-c$, $\text{ker}(\partial_2)=0$2, $\text{Im}(\partial_1)=0$, and $\text{ker}(\partial_1)=C_1(X)$, and so it follows that

$H_2(X)=\text{ker}(\partial_2)/\text{Im}(\partial_3)=0$,

$H_1(X)=\text{ker}(\partial_1)/\text{Im}(\partial_2)=C_1(X)/\langle a+b-c\rangle\cong\langle a,b\rangle=\mathbb{Z}^2$,

and

$H_0(X)=\text{ker}(\partial_0)/\text{Im}(\partial_1)=C_0(X)/0\cong\mathbb{Z}$.

$\square$

1. Image generated from LaTeX here: https://www.writelatex.com/read/tnhkhhzscfvj.

2. A typical element in $C_2(X)$ has the form $nF$, $n\in\mathbb{Z}$, and so $\partial_2:nF\mapsto n(a+b-c)$. In particular, $nF\in\text{ker}(\partial_2)$ if and only if $n(a+b-c)=0$, i.e. if and only if $n=0$.