Euler characteristic: dependence on coefficients Let $X$ be a finite CW complex and $\chi(X)$ its Euler characteristic (defined using integer coefficients). When is it true that $\chi(X)=\sum (-1)^i \dim H_i(X;F)$, where $F$ is a field?
I thought it would be true for all fields, but I noticed that for $X$ the Klein bottle and $F=\mathbb{Z}/2$ this is false! In fact Bredon (Geometry and topology) claims it to be true, but it isn't..
 A: There is also a way showing the independence of $\chi (X)$ without using the universal coefficient theorem, but using cellular homology:
Let $a_n$ denote the number of cells of $X$ in dimension $n$. Then we get the cellular chain complex 
$$\ldots \longrightarrow F^{a_k} \longrightarrow F^{a_{k-1}} \longrightarrow \ldots $$
Using that the Euler characteristic of a chain complex equals that of its Homology (which is proved using the dimension formula for vector spaces), we get 
$$\chi_F=\sum (-1)^i a_i$$ and 
$$\chi_F=\sum (-1)^i \dim H_i(X;F)$$
which is then independent of the field.
A: $\newcommand{\Z}{\mathbb{Z}}$The Klein bottle has the following integral homology groups: $$H_*(K; \Z) = (\Z, \Z \oplus \Z/2\Z, 0, 0, \dots).$$
This gives an Euler characteristic $\chi_\Z = 1-1 = 0$. Over $\Z/2\Z$, the universal coefficient theorem yields:
$$H_*(K; \Z/2\Z) = (\Z/2\Z, \Z/2\Z^2, \Z/2\Z, 0, 0, \dots)$$
and so the Euler characteristic is $\chi_{\Z/2\Z} = 1-2+1 = 0$. This agrees with the previous computation.

In general, suppose $X$ is a space with finite integral homology (i.e. a finite number of nonzero homology groups, and these groups are all finitely generated), for example a finite CW-complex. Then its Euler characteristic is well-defined. Let $n$ be the top dimension of nonvanishing homology.


*

*First suppose that $\newcommand{\F}{\mathbb{F}}\F$ is a field of characteristic zero. Let $b_k$ be the $k$th Betti number of $X$, ie.
$$H_k(X;\Z) = \Z^{b_k} \oplus T$$
where $T$ is the torsion subgroup. Then
$$\chi_\Z = b_0 - b_1 + b_2 - \dots = \sum_k (-1)^k b_k.$$
Now the universal coefficient theorem says $$H_k(X;\F) = H_k(X;\Z) \otimes \F \oplus \operatorname{Tor}(H_{k-1}(X;\Z), \F).$$
Since the field is of characteristic zero, the $\operatorname{Tor}$-term vanishes, and you're left with $H_k(X;\F) = \F^{b_k}$. It follows that $\chi_\F = \chi_\Z$.

*Suppose now that $\F$ is a field of characteristic $p$. Suppose also that $H_k(X; \Z) = \Z^{b_k} \oplus (\Z/p\Z)^{c^p_k} \oplus T^p_k$, where $T^p_k$ is the torsion part which is not $p$-torsion.
The universal coefficient theorem gives:
$$H_k(X; \F) = \begin{cases}
\F^{b_0 + c^p_0} & k = 0 \\
\F^{b_k + c^p_k + c^p_{k-1}} & 1 \le k \le n \\
\F^{c^p_n} & k = n+1
\end{cases}$$
(in this formula, I write $c^p_0$ knowing fully well that in fact $c^p_0 = 0$. It does not matter.)
Then the Euler characteristic becomes:
$$\chi_\F = (b_0 + c^p_0) - (b_1 + c^p_1 + c^p_0) + \dots + (-1)^n (b_n+c^p_n+c^p_{n-1}) + (-1)^{n+1}c^p_n$$
Each $c^p_k$ cancels with the one in the next factor, so all is left is
$$\chi_\F = b_0 - b_1 + \dots = \chi_\Z.$$

