I've got this question on Atiyah Commutative Algebra (first chapter, "Ideals and Quotient Rings" section).
Given a ring $A$ and a ideal $a \subseteq A$, there is a natural projection $\pi: A \rightarrow A/a$, $\, \, \, x \mapsto \pi(x) = x+a$, which is surjective. Then, given a ring $B$ and a homomorphism $f \in Hom(A,B)$, we have $A/a \cong f(A)$.
I've tried to manipulate $f$ and $\pi$ to get to this isomorphism $\phi$, which, in my guest, would be like $\phi (x) := f(\pi^{-1}(x))$. But, I've got stuck on the "bijection part" of the proof.
I would appreciate any comment on this little part of the book :)
Be careful with the map $\phi$ that you've defined since it actually isn't well defined! For instance, suppose that your ideal $a$ has two elements $x$ and $y$. Then, $x + a = y + a$, so $\pi^{-1}(x+a)$ is not a singleton (both $x$ and $y$ are in the pre-image). We also need to be careful about the ideal $a \subset A$ - it's not just any old ideal - in Atiyah's book, $a$ is stated to be the kernel of $f$!
Instead, define a map $g: A/a \to f(A)$ by sending $x + a \in A/a$ to $f(x)$. That is, $$g(x+a) := f(x)$$ We need to show this map is well defined. Indeed, if $y \in x+a$, then $y = x + m$ for some $m \in A$. Hence, $$f(y) = f(x+m) = f(x)+ f(m) = f(x)$$ as $m$ is in the kernel of $f$. So indeed, we will have $g(x+a) = g(y+a)$ and our map is well-defined. To show $g$ is injective, suppose $g(x+a) = g(y+a)$. Then, this says that $f(x) = f(y) \implies 0 = f(x)-f(y) = f(x-y)$. Hence, $x-y$ is in the kernel of $f$ (i.e. $x-y \in a$). This will in turn, because of the axioms of an ideal, imply that $x + a \subset y + a$. But I could run the argument again by noticing that $0 = f(y) - f(x) = f(y-x)$, so $y-x \in a \implies y+a \subset x+a$. Thus, $x+a = y+a$, and we have shown injectivity.
To show surjectivity, suppose $b \in f(A)$. Then, there exists $x \in A$ such that $f(x) = b$. Then, $g(x+a) = f(x) = b$.