Question

Let R be a ring and f : M −→ N a morphism of left R-modules. Show that:

c) K := {m ∈ M | f(m) = 0} satisfies the Universal Property of Kernels.

d) N/f(M) satisfies the Universal Property of Cokernels.

 

Answer 1

 

c)

Suppose that i : K M the inclusion.

Observe that f o i (x) = f(x) = 0 for all x in K.

Thus, foi is the zero map.

Now, suppose that K' is an R-module and g : K' M be a morphism such that f o g : K' N is the zero map. In other words, fog(x) = 0 for all x in K'. Thus, f(g(x)) = 0 for all x in K'. Hence, g(x) is in K for all x in K'.

Hence, g : K' K. Thus, g : K' M factors through K as g = i o g. Further if g' : K' K is another morphism such that g = i o g' , then observe that g(x) = i(g(x)) = i (g'(x)) = g'(x) for all x in K'. Hence, g : K' K is the unique morphism such that g : K' M factors through K as g = i o g.

Hence, K satisfies the Universal Property of Kernels.

d)

Consider the canonical projection π : N N/f(M) given by, π(n) = f(M) + n for all n in N.

Observe that π o f : M N/f(M) is the zero map. This is because, π o f(m) = f(M) + f(m) = f(M) + 0 (which is the identity of N/f(M)).

Further if Q is an R-module and π' : N ​​ Q is a morphism such that π' o f is the zero map, then f(M) is contained in ker π'. Thus, π' factors through N/f(M) via a unique morphism q : N/f(M) Q such that π' = q o π.