composition of maps in exact sequence is the 0 map?

Question

If I have an exact sequence of modules:

$0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0$

why do I have $g \circ f = 0$?

I am not even sure this is correct, but when I look at the examples, it seems correct :

$0 \rightarrow \mathbb{Z} \xrightarrow{i} B \xrightarrow{\pi} \mathbb{R}/\mathbb{Z} \rightarrow 0$

$0 \rightarrow Ker(T) \xrightarrow{i} A \xrightarrow{T} Im(T) \rightarrow 0$

So, is it true in general? Why?

Answer 1

By definition, an exact sequence is such that the image of one of the map is equal to the kernel of the following.

Here, it means that $\mathrm{Im}(f)=\mathrm{Ker}(g)$. Of course, this implies that $g \circ f = 0$ (because for every $x \in A$, $f(x) \in \mathrm{Im}(f)$ so $f(x) \in \mathrm{Ker}(g)$ so $g(f(x))=0$).