Second Homology Group of the Tetrahedron

Question

I've been computing the homologygroups of the Tetrahedron T with the "usual" maps, yielding the sequence $0 \rightarrow \mathbb{Z}^4 \rightarrow \mathbb{Z}^6 \rightarrow \mathbb{Z}^4 \rightarrow 0$. However, when I computed the matrices of the maps I got the result that $H_2(T)=\mathbb{Z}$. But shouldn't I get the result $H_2(T)= \{0\}$, since T is homotopically equivalent to a point (since it is a star domain.)?

Answer 1

Given what you indicate it is correct to obtain $H_2(T) = \mathbb{Z}$.

The dimensions in your sequence correspond from left to right to respectively the faces, the edges and the vertices, but if you want to compute the solid tetrahedron you need to add on the left one more term to take into account the interior of your tetrahedron.

Your computation here is for the boundary of the tetrahedron, and this shape is equivalent to a sphere.