Let R and R' be rings and let N and N' be ideals of R and R' respectively.
Let be a homomorphism of R into R'.
Show that induces a natural homomorphism if
well, what do we have to work with?
we are given the homomorphism φ:R-->R', the ideal N of R and the ideal N' of R', and that φ(N) is a subset of N'.
so the natural thing to do is define: φ*(r+N) = φ(r)+N'.
whenEVER you define things on cosets, it is imperative that you verify that the definition depends ONLY on the coset r+N, and not on "r".
so we must check that if r'+N = r+N, that φ*(r+N) = φ*(r'+N).
if r+N = r'+N, this means r-r' is in N. since φ maps N inside N', φ(r-r') is in N'. since φ is a homomorphism, φ(r-r') = φ(r)-φ(r').
so we have φ(r)-φ(r') is in N', hence φ(r)+N' = φ(r')+N', that is: φ*(r+N) = φ*(r'+N), as desired.
now all that is left to do is verify that φ* is a ring homomorphism. you can do this.