I am reading the proof for the correspondence theorem in group theory:
$\phi$ is a surjective homomorphism $G_1 \longrightarrow G_2$, then $\phi$ is a bijection between A={subgroups of $G_1$ contains the kernal($\phi$)} and B={subgroups of $G_2$}
I know there are couple ways to state this theorem (subgroups of the $G/S$ correspond to the subgroups of $G$ containing $S$), but this is the version I am working on
(I already proved that $\phi(A) \in B$ domain A codomain B and $\phi^{-1}(B) \in A$)
Question: Generally when I want to prove something is a bijection, I show that its one to one and onto: $$f(x)=f(y) \implies x=y$$ $$ \forall x \in B \implies f(y) = x, y\in A$$ Here I can show $\phi$ is one to one by proving that preimage is unique: $$\phi^{-1}(\phi(x)) = x, x \in A$$ Do I need to worry about onto? my notes suggest that I prove: $$\phi(\phi^{-1}(y)) = y, y \in B$$ But I already showed that $\phi^{-1}(y) \in A$ is that not enough? am I missing something in my proof?
You may know $\phi^{-1}(y)\in A$, but does that imply $\phi(\phi^{-1}(y))=y?$
In general, suppose you have a function $f: A \rightarrow B$ between some sets $A$ and $B$. Another way to say $f$ is bijective, is to say there is a function $g: B \rightarrow A$ such that $g\circ f (a) = a$ for all $a\in A$, and $f\circ g (b)=b$ for all $b\in B$. The condition $g\circ f(a)=a$ is equivalent to $f$ being one-to-one, and $f\circ g(b)=b$ is equivalent to $f$ being onto.
Here you have $f=\phi$, $g=\phi^{-1}$. You have shown $g\circ f(a)=a$, but not $f\circ g(b)=b$. Showing $\phi^{-1}(b)\in A$ just says $\phi^{-1}$ is a function from $B$ to $A$.