# Thread: commutative diagram, module homomorphisms

1. ## commutative diagram, module homomorphisms

Suppose is a commutative diagram of $\displaystyle R$-modules and $\displaystyle R$-module homomorphisms.
(a) Suppose that $\displaystyle \phi$ is injective. Prove that the map $\displaystyle \phi_0 : \text{Ker } f \rightarrow \text{Ker } g$ is injective.

(b) Suppose that $\displaystyle \psi$ is surjective. Prove that the map $\displaystyle \bar{\psi} : \text{Coker } f \rightarrow \text{Coker } g$ is surjective.

2. Originally Posted by fermatprime371
Suppose is a commutative diagram of $\displaystyle R$-modules and $\displaystyle R$-module homomorphisms.
(a) Suppose that $\displaystyle \phi$ is injective. Prove that the map $\displaystyle \phi_0 : \text{Ker } f \rightarrow \text{Ker } g$ is injective.

(b) Suppose that $\displaystyle \psi$ is surjective. Prove that the map $\displaystyle \bar{\psi} : \text{Coker } f \rightarrow \text{Coker } g$ is surjective.
this is trivial! the only thing which you need to notice is the definition of the maps $\displaystyle \phi_0, \ \bar{\psi},$ which are as follows: $\displaystyle \forall a \in \ker f: \ \phi_0(a)=\phi(a)$ and $\displaystyle \forall a' \in A': \ \bar{\psi}(a'+ \text{Im} f)=\psi(a')+\text{Im} g.$

these maps are well-defined because of the commutativity of your diagram: if $\displaystyle a \in \ker f,$ then $\displaystyle g \phi(a)=\psi f(a)=0.$ thus $\displaystyle \phi(a) \in \ker g.$ also if $\displaystyle a_1' + \in \text{Im}f=a_2'+\text{Im}f,$ then $\displaystyle a_1'-a_2' \in \text{Im}f.$

thus $\displaystyle \exists a \in A : \ f(a)=a_1' - a_2 '.$ hence: $\displaystyle \psi(a_1')-\psi(a_2')=\psi(a_1'-a_2')=\psi f(a)=g \phi(a) \in \text{Im}g.$ therefore $\displaystyle \bar{\psi}(a_1' + \text{Im}f)=\bar{\psi}(a_2' + \text{Im}f).$ so the maps $\displaystyle \phi_0$ and $\displaystyle \bar{\psi}$ are well-defined. suppose $\displaystyle \phi$

is inective and $\displaystyle \phi_0(a_1)=\phi_0(a_2).$ so from the definition of $\displaystyle \phi_0$ we'll get $\displaystyle \phi(a_1)=\phi(a_2).$ thus $\displaystyle a_1=a_2,$ because $\displaystyle \phi$ is injective. if $\displaystyle \psi$ is surjective and $\displaystyle b' \in B',$ then $\displaystyle \psi(a')=b',$ for some $\displaystyle a' \in A'.$

thus: $\displaystyle \bar{\psi}(a'+ \text{Im}f)=\psi(a')+\text{Im}g=b' + \text{Im}g. \ \Box$