Proof
Put I = Ker(f). This is an ideal of R. We shall prove the following two assertions.
There is a morphism f' : R/I -> S
such that, for each x 
R,
f(x) = f'(x + I).
The morphism f' is injective.
Since f'(R) = f(R) ,
it is then immediate that
the resulting morphism R/I -> f(R),
being injective and by choice of codomain surjective, is bijective and hence an isomorphism.
The map f' is determined by the requirement
f'(x + I) = f(x).
It needs to be verified that f' is well defined.
For, if x + I = y + I, then
x - y I = Ker(f),
so
f(x - y) = 0, whence, as f is a morphism,
f(x) = f(y).
Thus, indeed, the definition of f' does not depend on the choice of
y x + I.
It is easy to see that f is a morphism.
By an earlier theorem,
it suffices to prove that Ker(f') = {0}.
Suppose, to this end, that x + I
Ker(f').
Then f(x) = 0, and so x Ker(f),
which is I.
Consequently, x + I = I = 0 R/I.
So indeed, Ker(f') = {0}.
Ker(f').
Then f(x) = 0, and so x Ker(f),
which is I.
Consequently, x + I = I = 0 R/I.
So indeed, Ker(f') = {0}.