- #1

- 90

- 2

If you look at the definition they give for ring homomorphism, they require only that addition and multiplication is preserved over the homomorphism (and not that it maps 1 to 1'). Then, under the 'Properties' section, the third bullet down claims:

Don't you need to explicitly require that the homomorphism maps 1 to 1' in order for that statement to be true? Or is there someway to deduce this without specifying that requirement?If a has a multiplicative inverse in R, then f(a) has a multiplicative inverse in S and we have f(a^{−1}) = (f(a))^{−1}. Therefore, f induces a group homomorphism from the (multiplicative) group of units of R to the (multiplicative) group of units of S.