[Math] Euler-Poincaré characteristic and homology

abelian-groupshomological-algebra

The problem

$\DeclareMathOperator{rk}{\text{rk}}$
$\DeclareMathOperator{im}{\text{im}}$
Let
$$C = ( C_n \overset{\partial_n}\to C_{n-1} \overset{\partial_{n-1}}\to \dots \overset{\partial_2}\to C_1 \overset{\partial_1}\to C_0 ) $$ be a chain complex of finitely-generated free abelian groups, and define its Euler-Poincaré characteristic by the formula
$$ \chi(C) = \sum_{i=0}^n (-1)^{i} \,\text{rk}(C_i) $$
It's straightforward to prove that $\chi(D)$ vanishes for a short exact sequence of abelian groups. Now I'm asked to use that result to show that
$$ \chi(C) = \chi(H(C)), $$
where $\chi(H(C))$ is the homology complex of $C$, in the natural sense.

My Solution

Define $Z_i = \ker(\partial_i)$ and $B_{i-1} = \im(\partial_i)$.
In our original chain complex $C$ we have levelwise short exact sequences
$$ 0\to Z_i \to C_i \to B_{i-1} \to 0, $$

while in the homology complex we have levelwise short exact sequences of the form
$$ 0\to B_i \to Z_i \to H_i \to 0. $$
Since both $B_{i-1}$ and $H_i$ are free abelian group, so equivalently free $\mathbb{Z}$-modules, they are in particular projective modules. Hence the above sequences split, giving
$$C_i = B_{i-1}\oplus Z_{i} \quad \implies \quad\rk(C_i) = \rk(B_{i-1}) – \rk(Z_{i})$$
and
$$Z_i = B_i\oplus H_i \quad \implies \quad\rk(H_i) = \rk(Z_i) – \rk(B_i).$$

Thus keeping in mind that $B_{-1} = B_{n} = 0$, we find
$$
\chi(C) – \chi(H(C)) = \sum_{i=0}^n (-1)^{i} \,\text{rk}(C_i) – \text{rk}(H_i) = (-1)^n B_n – B_{-1} = 0 $$

Questions

  1. How can one extend this to a chain complex made of non necessarily free abelian groups?

  2. Do you agree with my solution? Do you have any remark to add?

  3. Are there any problems generalizing this to free $R$-module, for a ring $R$ with the Invariant Basis Number property?

Best Answer

Let $P$ be a Euler-Poincaré map. We have the so called "fundamental" short exact sequences

$$0\longrightarrow B_n\longrightarrow Z_n \longrightarrow H_n \longrightarrow 0\\ 0 \longrightarrow Z_n \longrightarrow C_n \longrightarrow B_{n+1} \longrightarrow 0$$ so that $P(C_n)=P(Z_n)+P(B_{n+1})$ and $ P(Z_n)=P(B_n)+P(H_n)$ Then $$\begin{align}\chi(C)&= \sum (-1)^i P(C_i)\\&=\sum (-1)^i P(Z_i) + \sum (-1)^i P(B_{i+1})\\&=\sum (-1)^i P(Z_i) - \sum (-1)^i P(B_i)\\&=\sum (-1)^i P(H_i)=\chi(H(C))\end{align}$$

This proves for example that weakly equivalent complexes have equal characterisitc. In partiular homotopy equivalent complexes have equal characterisitc, so indeed this characteristic is an homotopy invariant.

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