The question is:

Let f : R ! S be a homomorphism. Let
T = { f(r) : r 2 R}.
T is called the image of f. Show that T is a subring of S.
Hint: In this problem, you can use the following facts:
f(0R) = 0S,f(−x) = −f(x) for x 2 R, and f(x − y) = f(x) − f(y) for x, y 2 R. These facts were stated in class for isomorphisms only but the same properties (with the same proofs) hold
for homomorphisms.

Thanks very much.

2. Originally Posted by suedenation
The question is:

Let f : R ! S be a homomorphism. Let
T = { f(r) : r 2 R}.
T is called the image of f. Show that T is a subring of S.
Hint: In this problem, you can use the following facts:
f(0R) = 0S,f(−x) = −f(x) for x 2 R, and f(x − y) = f(x) − f(y) for x, y 2 R. These facts were stated in class for isomorphisms only but the same properties (with the same proofs) hold
for homomorphisms.

Thanks very much.
You forgot to mention it, but R and S must be rings.

T is, of course, a subset of S.

By definition of the homomorphism, f(a+b)=f(a)+f(b) and f(a*b)=f(a)*f(b).

If a set T is a ring, it must have the following properties:
1. a+b=b+a
2. (a+b)+c=a+(b+c)
3. An identity (0) for addition exists.
4. An additive inverse exists for all elements of T.
5. (a*b)*c=a*(b*c)
6. An identity (1) for multiplication exists.
7. * distributes over +.

You can use the homomorphism to assist you. For example, the first property:
f(a+b)=f(a)+f(b)
f(b+a)=f(b)+f(a)
Since f(a+b)=f(b+a) we have that f(a)+f(b)=f(b)+f(a). So addition in T is commutative.

The second property follows the same pattern.
The third property: f(a+0)=f(a)+f(0)=f(a), so f(0)=0T exists.
The fourth property: f(a+-a)=f(a)+f(-a)=f(0)=0T, so f(-a)=-f(a) exists.
The fifth property:
f((a*b)*c)=(f(a)*f(b))*f(c)
f(a*(b*c))=f(a)*(f(b)*f(c))
and the two of these are equal.

The sixth property f(a*1)=f(a)*f(1)=f(a) so f(1)=1T exists.
The seventh property: f(a*(b+c))=f(a)*(f(b)+f(c)) etc.

And, of course, you should put in a statement that 0T and 1T correspond to 0S and 1S respectively, which is easy to show.

-Dan