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:

Hatcher 2.1.4

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)=02, \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 $n=0$.

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