# Snake lemma for $R$-modules. Help with $\ker$ maps, not connecting hom.

In Construction of the maps they say that the construction of the kernel/cokernel maps comes from the commutativity of the diagram:

$$\require{AMScd} \begin{CD} & & A @>{f}>> B @>{g}>>C @>>> 0 \\ & @VV{a}V & @VV{b}V & @VV{c}V \\ 0 @>>> A' @>{f'}>> B' @>{g'}>> C' \\ \end{CD}$$

Define $\hat{f} = f\mid_{\ker a}$. I want to show that $\hat{f}(A) \subset \ker b$. So that $\ker a \xrightarrow{\hat{f}} \ker b$ exists.

Let $x \in \ker{a}$, then $b\hat{f}(x) = bf(x) = f'a(x) = 0$ so that $f(x) \in \ker b$. Next, if $\hat{g}$ is defined similarly, then I want to show exactness or $\ker \hat{g} = \text{im} \hat{f}$. First, $\ker \hat{g} \supset \text{im} \hat{f}$ or $\hat{g}\hat{f} = 0$.

I don't see how to show this.

We know that $\ker g = \text{im} f$ so that $\ker g \cap \ker b = \text{im}f \cap \ker b \supset \text{im} \hat{f}$...

## 1 Answer

The inclusion of $\mathrm{ker} c$ into $C$ is a monomorphism, so $\hat g\hat f=0$ because $gf=0$.