= Solution
The <universal coefficient theorem for homology> and flatness of $\mathbb Q$ give
$$
H_i(X;\mathbb Q)
\cong H_i(X;\mathbb Z)\otimes_\mathbb Z\mathbb Q.
$$
Thus, if
$$
H_i(X;\mathbb Z)\cong\mathbb Z^{b_i}\oplus T_i
$$
with $T_i$ torsion, then $H_i(X;\mathbb Q)\cong\mathbb Q^{b_i}$.
For a finite triangulable space, its <Euler characteristic> is
$$
\chi(X)=\sum_i(-1)^i\dim_\mathbb QH_i(X;\mathbb Q).
$$
If $K$ is a finite triangulation and $f_i$ is its number of $i$-simplices, then the <Euler-Poincare formula> is
$$
\chi(X)=\sum_i(-1)^if_i.
$$
To prove it over $\mathbb Q$, write $Z_i=\ker\partial_i$ and $B_i=\operatorname{im}\partial_{i+1}$. Rank-nullity and $H_i=Z_i/B_i$ give
$$
\dim C_i
=\dim Z_i+\dim B_{i-1}
=\dim H_i+\dim B_i+\dim B_{i-1}.
$$
After multiplying by $(-1)^i$ and summing, the two boundary sums cancel, leaving the claimed equality.
The standard simplex $\Delta^3$ is contractible, so
$$
H_i(\Delta^3)\cong
\begin{cases}
\mathbb Z,&i=0,\\
0,&i>0.
\end{cases}
$$
Back to article page
