If we have two rings with Unity $A,B$ and a epimorphism $\phi$. Then $\phi(1_A)=\phi(1_B)$.
I have difficulties to understand this proof:
$\phi \text{ surj.}\Rightarrow \forall b\in B \exists a\in A:\phi(a)=b\Rightarrow \forall b\in B\exists a\in A: \phi(1)\phi(a)=\phi(1)b\Rightarrow \forall b\in B:b=\phi(1)b$
and similarily the rightinverse.
I don't understand what I don't understand but the fact that $b$ can have more than one preimages bothers me. Can somebody explain me why we just have to look at one preimage and why we didn't Mention the other preimages in the above proof?Please?
Presumably, you're referring to this answer to your previous question. It's difficult to address a concern that you can't articulate. Let's run over the proof again:
Suppose $\phi : A \to B$ is a homomorphism between rings, $A$ has a multiplicative identity $1_A$, and $\phi$ is surjective. We wish to show $\phi(1_A)$ is a multiplicative identity of $B$.
To prove this, consider an element $b \in B$. Since $\phi$ is surjective, there exists (at least one) $a \in A$ such that $\phi(a) = b$. Therefore, $$b\phi(1_A) = \phi(a)\phi(1_A) = \phi(a1_A) = \phi(a) = b,$$ and similarly, $$\phi(1_A)b = b.$$
As I think I understand your objection, your issue is that there may be several possible values of $a \in A$ such that $\phi(a) = b$. The proof involves selecting one, without any further information, and running with it. What if you select the wrong value of $a$? Is the proof not just falsely pretending that there's only one value of $a \in A$ satisfying $\phi(a) = b$?
Well, no, not really. There may be multiple (infinitely many, even) values of $a$ such that $\phi(a) = b$, but choosing any one will do the job fine. At the end, no matter which $a$ we selected to do the job, the result at the end of the calculation is always $\phi(a)$, and the only criteria we asked of $a$ is that $\phi(a)$ be the same thing as $b$.
To demonstrate this, let's consider a concrete example. Consider the ring $\Bbb{R}^2 = \Bbb{R} \times \Bbb{R}$, meaning that we equip with the usual componentwise addition, as well as a componentwise multiplication: $$(a, b)(c, d) = (ac, bd).$$ Then $(1, 1)$ is a multiplicative identity on $\Bbb{R}^2$. Consider the ring homomorphism: $$\phi : \Bbb{R}^2 \to \Bbb{R} : (x, y) \mapsto x.$$ This is surjective, as given any $x \in \Bbb{R}$, the point $(x, 0) \mapsto x$. It's also far from injective, as $(x, 1), (x, 7), (x, \pi),$ and infinitely many others all map to $x$ as well.
Let's roll through the argument. Suppose we have $b \in \Bbb{R}$. Then, as we know $\phi$ is surjective; we indeed have a recipe for preimages of $b$. As such, we will make the arbitrary choice of $a = (b, 0)$. Then, $$b \phi(1, 1) = \phi(b, 0)\phi(1, 1) = \phi((b, 0)(1, 1)) = \phi(b, 0) = b.$$ That worked exactly as the proof advertised. What about with a different preimage than $(b, 0)$? Try again with $a = (b , 1)$: $$b \phi(1, 1) = \phi(b, 1)\phi(1, 1) = \phi((b, 1)(1, 1)) = \phi(b, 1) = b.$$ Again, it works as advertised. It doesn't matter which $a$ we choose, so long as $\phi(a) = b$, and this proof will work. It means we have to perform the $\phi(a)\phi(1_A) = \phi(a1_A)$ step without knowing specifically what $a$ is, but fortunately, this works for all elements of $A$, so we don't need to have perfect knowledge of $a$. Again, all we need from $a$ in the end is $\phi(a) = b$, and having many options for choosing $a$ is no obstacle.