Provided is a homomorphism of short exact sequences:
$$ \begin{array} & 0 & \rightarrow & F_1 & \overset{\varphi_1}{\rightarrow} & E_1 & \overset{\psi_1}{\rightarrow} & G_1 & \rightarrow& 0 \\ & & \varrho\downarrow & & \sigma\downarrow & & \tau\downarrow & & \\ 0 & \rightarrow & F_2 & \overset{\varphi_2}{\rightarrow} & E_2 & \overset{\psi_2}{\rightarrow} & G_2 & \rightarrow& 0 \\ \end{array} $$
I am asked to show that
$$\varphi_1(\text{ker}\varrho)=\text{ker}\sigma$$
but I do not know how to start. Could you give me a suggestion?
Let $b \in \varphi_1 (\ker \rho)$, then existis $a\in \ker \rho$ such that $b=\varphi_1(a)$, then $\sigma(b)= \sigma(\varphi_1(a))$, now $\sigma(b)= \varphi_2(\rho(a))$. So $\sigma(b)=0$, then $b \in \ker(\sigma)$