TITLE: "Almost" ring homomorphism
QUESTION [6 upvotes]: This is an exercise out of Herstein which seems pretty straightforward but is eluding me.
Let $R,R'$ be rings and let $\phi:R\to R'$ be a mapping such that, for every $x,y\in R$:
$${\rm 1.}\qquad\phi(x+y)=\phi(x)+\phi(y)$$
$${\rm 2.}\qquad\phi(xy)=\phi(x)\phi(y)\qquad{\rm or}\qquad\phi(xy)=\phi(y)\phi(x)$$
I'm to show that one of the conditions holds uniformly. That is, for all $x,y\in R$, $\phi(xy)=\phi(x)\phi(y)$ or, for all $x,y\in R$, $\phi(xy)=\phi(y)\phi(x)$. Of course, this doesn't exclude the possibility that both conditions hold uniformly (which happens when e.g. $R'$ is commutative).
Herstein hints to fix $a\in R$ and to consider the sets
$$W_a=\{x\in R\mid\phi(ax)=\phi(a)\phi(x)\},$$
$$V_a=\{x\in R\mid\phi(ax)=\phi(x)\phi(a)\}.$$
I have shown that one of $W_a$ or $V_a$ must be the entire ring $R$. I think I'm missing something simple to complete the argument. Any hints?
Edit
I forgot to mention that in my copy of the text, the condition to be shown is stated as "$\phi(xy)=\phi(x)\phi(y)$ or $\phi(x)=\phi(y)\phi(x)$". I am pretty confident that it's a typo, but maybe not?

REPLY [3 votes]: Fact. If a group $G$ is a union of two subgroups $G = G_1 \cup G_2$, then $G=G_1$ or $G=G_2$.
We use this two times. Let $a \in R$. Then $W_a$ and $V_a$ are additive subgroups of $R$ with union $R$. Hence, $W_a=R$ or $V_a=R$. Now, $\{a \in R : W_a=R\}$ and $\{a \in R : V_a=R\}$ are additive subgroups of $R$ with union $R$. Hence, $\{a \in R : W_a=R\}$, i.e. $\phi$ is a homomorphism, or $\{a \in R :V_a=R\}$, i.e. $\phi$ is an anti-homomorphism.