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Take $\varphi: R_1 \longrightarrow R_2$, and $f:G_1 \longrightarrow G_2$.

In order to prove $\varphi$ is a ring homomorphism, we must show $\varphi(1_{R_1})=1_{R_2}$ (in addition to other properties), but to prove $f$ is a group homomorphism it suffices just to show $f(g_1\cdot g_2) = f(g_1)\cdot f(g_2)$, with $f(e_{G_1})=e_{G_2}$ being implied - how is this so?

Referencing this post about why we need to specify $\varphi(1_{R_1})=1_{R_2}$, the zero map is brought up as an example of why we need this specification, since $0_{R2}$ might not be the identity element of $R_2$, but $\varphi(1_{R_1}\cdot r) = \varphi(1_{R_1})\cdot\varphi(r)= 0 \cdot 0$, so $0_{R_2} \textbf{will}$ be the identity element of $\varphi(R_1)$.

But doesn't this same reasoning apply to our group homomorphism $f$? The following proof from my textbook for why $f(e_{G_1}) = e_{G_2}$ seems like it should theoretically hold for rings...

$e\cdot g = g$

$f(e\cdot g) = f(g)$

$\Rightarrow f(e) \cdot f(g) = f(g)$

$\Rightarrow f(e)$ is the identity element of $G_2$

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    $\begingroup$ You can't cancel by $f(g)$ in a ring. $\endgroup$ Feb 28, 2022 at 3:49
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    $\begingroup$ It sounds like you're mixing up the category of rings [with multiplicative identity] and the category of rngs [which may not have a multiplicative identity]. There's a theory for both, and in the category of rngs, when $R_1$ has a multiplicative identity $1_{R_1}$, its image $f(1_{R_1}) \in R_2$ will always be the multiplicative identity of the rng $f(R_1)$ even when $R_2$ doesn't have a multiplicative identity. This is, in essence, what you seem to be saying. $\endgroup$ Feb 28, 2022 at 4:26

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The last implication assumes that $f(g)$ is invertible (since you only multiplied by $f(g)^{-1}$ to the right on both sides to get that $f(e)$ is the identity element), and this might not be the case for a ring, since there are ring elements that are not invertible. In the counterexample that you pointed out, as you said, $f$ could be the zero map and, in this case, $f(x)$ is not invertible, so the argument used for group homomorphism does not apply there.

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