Homology groups of a square

algebraic-topologyhomology-cohomologysimplicial-complex

Consider the unit square with a $\Delta$– complex structure obtained by the two $2$-simplices $[v_0,v_1,v_2], [v_2,v_0,v_3]$. Is $\Delta{}_0(X)\cong\mathbb{Z}^6$ or $\mathbb{Z}^4$ where $X$ is the space of the square.

My confusion lies with the diagonal 1-simplex $[v_0,v_2]$. Its clear there are 5 1 simplices, but are there 4 or 6 0-simplices? surely if there are 6, that contradicts the requirements for a $\Delta$-complex structure?

edit: $v_i$ are the $i^{th}$ corner of the square.

Best Answer

There are 4 0-simplices. You got the square by taking two triangles and identifying two vertices from each diagonal

Related Question