# Thread: induced sequence of quotients

1. ## induced sequence of quotients

Hi! Let $A$ a finitely generated free abelian group and $f: A \rightarrow A$ a homomorphism, then one can define a trace as usually as sum of the diagonal elements (of the matrix form of $f$).

If $A$ is not free, then $\frac{A}{T(A)}$ is free ( $T(A)$ denotes the torsion group) and one has an induced homomorphism $\tilde f: \frac{A}{T(A)} \rightarrow \frac{A}{T(A)}$. Therefore we can define the trace of $f$ as $tr(f):=tr(\tilde f: \frac{A}{T(A)} \rightarrow \frac{A}{T(A)})$.

Now my question:
Let $A, \, B, \, C$ finitely generated abelian groups and
$
0 \longrightarrow A \overset{\text{i}}{\longrightarrow} B \overset{\text{p}}{\longrightarrow} C \longrightarrow 0
$

$
0 \longrightarrow A \overset{\text{i}}{\longrightarrow} B \overset{\text{p}}{\longrightarrow} C \longrightarrow 0
$

a commuative diagramm with exact rows (there should be vertical arrows: $f: A \rightarrow A, \, g: B \rightarrow B, \, h: C \rightarrow C$, but i do not know how to produce this. I hope it is clear what i mean).

Why do we have $tr(g)=tr(f)+tr(h)$ in this case?
My problem is that the induced sequence between the quotients is not longer exact anymore, since we lose the property $ker(\tilde p) \subset im(\tilde i)$.

Does somebody have a clue how to prove this?

2. Well it works with a simple ruse: One just considers the induced sequence of tensor products with Q, then we have an exact sequence and the trace won't change.