How does the proof of the Second Isomorphism theorem of rings prove this theorem? The theorem is :

Let $I$ be an ideal of the ring $R$.
Then there is a one-to-one correspondence between the subrings of $R/I$ and the subrings of $R$ containing $I$, given as follows: to a subring $S$ of $R$ containing $I$ corresponds the subring $ S/I $ of $R/I$. Under this correspondence, ideals of $R/I$ correspond to ideals of $R$ containing $I$; and, if $J$ is an ideal of $R$ containing $I$, then
$(R/I)/(J/I) \cong R/J$

I understood everything in the proof, however they prove it as if it were an equivalence, and not a bijection, what am I missing?
The equivalence would be
$$ (\exists S)(S\leq R\ \land\ I\subset S\ \land\ A=S/I)\quad \iff\quad A\leq R/I.$$
But in the theorem they stated a function like this:
$$ \Phi : \{S:S \leq R \land I \subset S \} \mapsto \{A: A\leq R/I \} $$
Would be a bijection. I've seen this in many points of view, asked my colleagues, teacher, older colleagues and didn't get satisfied with any answer.
I tried to see it, the implication in this direction $\rightarrow$ prove that it's well defined and $\leftarrow$ proves that it's surjective, but where is injectivity?
Another way is, $\rightarrow $ proves that it's a function and $\leftarrow$ proves that it's inverse is also a function, so it's a bijection, but I'm not so conviced with this argument, could someone help me?
Can I prove bijections with equivalences?
 A: On its own, given a ring $R$ and an ideal $I\subset R$, the logical statement
$$S \leq R\quad \land\quad I \subset S\qquad \iff\qquad S/I \leq R/I,$$
doesn't quite make sense; if the left hand side is not satisfied, then the right hand side may be undefined. What is $S/I$ supposed to mean if $I\not\subset S$? Or worse, if $S\not\leq R$? (I take it that $S\leq R$ means that $S$ is a subring of $R$.)
As you suggest in the comments, you might instead prove the two equivalences
\begin{eqnarray*}
&&S \leq R\ \land\ I \subset S\ &\Longrightarrow&\qquad S/I \leq R/I,\\
(\exists S)(&&S\leq R\ \land\ I\subset S\ \land\ A=S/I)\qquad &\Longleftarrow&\qquad \hspace{9pt}A\leq R/I.
\end{eqnarray*}
The first is equivalent to $\Phi$ being well-defined, the second is equivalent to $\Phi$ being surjective. Neither shows that $\Phi$ is injective; this is equivalent to the $S$ in the second implication being unique. So the claim that $\Phi$ is a bijection is equivalent to the implication
$$(\exists!S)(S\leq R\ \land\ I\subset S\ \land\ A=S/I)\qquad \Longleftarrow\qquad A\leq R/I.$$
