Let $M \ge N \ge P$ be R-modules. Prove that there exist natural (not depending on choices) R-homomorphisms $N/P \to M/P$ and $M/P \to M/N$ for which the sequence $0 \to N/P \to M/P \to M/N \to 0$ is exact.
I am having difficulty with just about everything to prove this--even what exactly it is asking. I would be assuming that the sequence given is exact and then proving that $N/P \to M/P$ and $M/P \to M/N$ are homomorphisms?
How does one go about "not making choices"?
Also, could you give me a hint on how to prove this? If I assume that it is exact, then I would have the $f: N/P \to M/P$ is injective, and $g:M/P \to M/N$ is surjective.
Hint;
Define the map $f: N/P \to M/P$ to be $f(n+P)=n+P$ this is an injective $R-$ module homomorphism, and the map $g: M/P \to M/N$ to be $g(m+P)=m+N$ this is a surjective $R-$ module homomorphism.
Or;
From $N/P\le M/P$ we get the following exact sequence $$0\to N/P \to M/P \to (M/P)/(N/P)\simeq M/N\to0$$