# Find the homology group of a certain simplicial complex

There is an exercise on my textbook (translated from Chinese):

Calculate the homology group of the simplicial complex $K$ corresponding to the following graph

It is difficult for me to find the $n$-simplex in $K$. Denote $K_n$ as the set of $n$-simplex of $K$, then it is clear that $K_n$ is empty for $n \neq 0,1,2$. For $K_0$, $K_1$ and $K_2$, I considered three possibilities:

1. $K_2 = \{ [P_0P_1P_4], [P_1P_2P_4], [P_2P_3P_4],[P_0P_1P_2],[P_1P_2P_3] \}$; $K_1 =\{ [P_0P_1], [P_1P_2],[P_2P_3],[P_0P_2],[P_0P_4], [P_2P_4],[P_1P_3],[P_1P_4],[P_3P_4] \}$; $K_0 = \{ [P_0],[P_1],[P_2],[P_3],[P_4] \}$.

2. $K_2 = \{ [P_0P_1P_4], [P_1P_2P_4], [P_2P_3P_4] \}$; $K_1 =\{ [P_0P_1], [P_1P_2],[P_2P_3],[P_0P_4], [P_2P_4],[P_1P_4],[P_3P_4] \}$; $K_0 = \{ [P_0],[P_1],[P_2],[P_3],[P_4] \}$.

3. $K_2 = \{ P_0P_1P_2],[P_1P_2P_3] \}$; $K_1 =\{ [P_0P_1], [P_1P_2],[P_2P_3],[P_0P_2],[P_1P_3] \}$; $K_0 = \{ [P_0],[P_1],[P_2],[P_3] \}$.

All of these had Euler characteristic $1$.

But the answer given in the book says $H_1(K) \cong R^3$, $H_0(K) \cong R$, $H_n(K) = 0$ for other $n$. The Euler characteristic is not $1$, so I was wrong.

Would you please tell me what are the $n$-simplex?

Many thanks!

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I think you are meant to assume that the triangles are "hollow". In other words, the complex has only 0-cells and 1-cells. I also think that the one-cells are exactly the lines that have been drawn in the figure; there are no line passing "on top of each other". So $K_1$ as in your second suggestion.
Just wondering, even then, how do you get $H_1(K)=R^3$? –  Haikal Yeo Apr 30 '13 at 12:43
the picture is a wedge of three circles. each circle contributes a dimension to $H_1$. –  Dan Petersen May 2 '13 at 7:11