# Show that $\ker\sigma \subset \varphi_1(\ker\rho)$ and $\operatorname{im}\tau \subset \psi_2(\operatorname{im}\sigma)$.

Given the homomorphism of short exact sequences, I must show that $$\ker\sigma \subset \varphi_{1}(\ker\rho)$$ and $$\operatorname{im}\tau \subset \psi_{2}(\operatorname{im}\sigma)$$. For the first part, let $$y \in \ker\sigma$$, that is, $$\sigma(y) = 0$$. (Is it correct to say that $$\psi_1(y) \in \psi_1 (\ker \sigma)$$ ?). I must find a $$x \in \ker \rho$$ such that $$\varphi_1(x) = y$$

On the other hand, I have the following question: If $$\tau$$ is surjective, can I say that $$\sigma$$ is surjective?

• The first part is false. If all the vertical morphisms are zero, you need to show that $E_1$ is a subset of the image of $\varphi_1$. It’s the reverse inclusion instead. Jun 14, 2020 at 21:43

As stated and illustrated in the comments by Mindlack the first part is clearly wrong. Consider $$\rho=\sigma=\tau=0$$ the trivial homomorphism. Then $$\ker\rho=F_1$$ and $$\ker\sigma= E_1$$ and you are asked to show that $$E_1\subset\varphi_1(F_1)$$ which is only the case for $$\varphi_1$$ being surjective. Pick your favorite non-surjective $$\varphi_1$$ and see that this cannot work. However, the reverse inclusion holds.

To see this, consider $$x\in\ker\rho$$. Then by commutativity $$(\varphi_2\circ\rho)(x)=(\sigma\circ\varphi_1)(x)=0$$ and thus $$\varphi_1(\ker\rho)\subset\ker\sigma$$. The second part is correct. Pick $$\tau(x)\in{\rm im}\,\tau$$ for some $$x\in G_1$$. By exactness, $$\psi_1$$ is surjective and therefore there is some $$y\in E_1$$ such that $$\psi_1(y)=x$$. Then $$(\psi_2\circ\sigma)(y)=(\tau\circ\psi_1)(y)=\tau(x)$$ and thus $${\rm im}\,\tau\subset\psi_2({\rm im}\,\sigma)$$.

Arguments like this are commonly known as proofs by diagram chase.

As $$y\in\ker\rho$$ it is correct to say $$\psi_1(y)\in\psi_1(\ker\rho)$$ as this is nothing else than applying $$\psi_1$$.

EDIT: Concerning the question below: consider

where the arrows are the respective inclusions and projections from the direct sums (and the identity for the rightmost column).$$\tau$$, in your notation, is surjective but $$\sigma$$ clearly not.

• OK thanks. I have the following question: If $\tau$ is surjective, can I say that $\sigma$ is surjective? Jun 15, 2020 at 2:46
• @JhöśëElijäh Not necessarily. Jun 15, 2020 at 2:50
• I have seen that if $\psi_ {2}$ is injective, it is true. But I am looking for a counterexample for the case where this linear application is not injective. Could you give me a suggestion? Jun 15, 2020 at 3:00
• @JhöśëElijäh Actually, it works the other way around, that is $\sigma$ being surjective implies $\tau$ being surjective. Do you mean 'linear map' with 'linear application'; otherwise I'm not familiar with this term. Jun 15, 2020 at 3:16
• Yes, I mean linear map. Excuse me. Jun 15, 2020 at 3:21