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Math Help - commutative diagram, module homomorphisms

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    commutative diagram, module homomorphisms

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

    (b) Suppose that \psi is surjective. Prove that the map \bar{\psi} : \text{Coker } f \rightarrow \text{Coker } g is surjective.
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  2. #2
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    Quote Originally Posted by fermatprime371 View Post
    Suppose is a commutative diagram of R-modules and R-module homomorphisms.
    (a) Suppose that \phi is injective. Prove that the map \phi_0 : \text{Ker } f \rightarrow \text{Ker } g is injective.

    (b) Suppose that \psi is surjective. Prove that the map \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 \phi_0, \ \bar{\psi}, which are as follows: \forall a \in \ker f: \ \phi_0(a)=\phi(a) and \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 a \in \ker f, then g \phi(a)=\psi f(a)=0. thus \phi(a) \in \ker g. also if a_1' + \in \text{Im}f=a_2'+\text{Im}f, then a_1'-a_2' \in \text{Im}f.

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

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

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