I know the following theorem:
For groups $(G_1,\circ)$ and $(G_2,\star)$, and for the homomorphism $\phi:(G_1,\circ) \rightarrow (G_2,\star)$:
If $H \leq G_1$, then $\phi(H) \leq G_2$
I wonder whether or not the (somewhat-of-a) converse was true?
In the event the $\phi$ is an isomorphism, then this is straightforward. However, what if $\phi$ is not injective but is surjective? Consider the notation $\phi^{-1}$, which simply denotes the inverse relation (not function) of $\phi$. Can we show that if $J \leq G_2$, then $\phi^{-1}(J) \leq G_1$? If not, might someone supply a counter example?
The motivation for this question stems from the standard pedagogical approach to teach the theorem that the kernel of $\phi$ is a subgroup of $G_1$...which seems to follow immediately if the above stated proposition is true.
Proposition. Let $f\colon G\to K$ be a group homomorphism. If $H\leq K$, then $f^{-1}(H)\leq G$.
Proof. Since $e_K\in H$ and $f(e_G)=e_K$, we have $e_G\in f^{-1}(H)$. So the latter is nonempty.
Let $x,y\in f^{-1}(H)$. We want to show that $xy^{-1}\in f^{-1}(H)$.
Since $x,y\in f^{-1}(H)$, then $f(x),f(y)\in H$. Since $H$ is a subgroup, it follows that $f(x)f(y)^{-1}\in H$. But $$f(x)f(y)^{-1} = f(x)f(y^{-1}) = f(xy^{-1}).$$ Thus, $f(xy^{-1})\in H$, so $xy^{-1}\in f^{-1}(H)$. We have thus shown that if $x,y\in f^{-1}(H)$, then $xy^{-1}\in f^{-1}(H)$.
This proves that $f^{-1}(H)\leq G$, as desired. $\Box$
Note, however, that this follows from the usual isomorphism theorems as well. The correspondence theorem tells you that this holds for surjective morphisms. Now note that $f^{-1}(H) = f^{-1}(H\cap f(G))$, so we can restrict attention to $f(G)$ and its subgroups... and then this is just part of the correspondence theorem.
On the other hand, this isn't really the "converse" of the theorem. The converse would say that if $f(X)$ is a subgroup of $K$, then $X$ is a subgroup of $G$. However, this is false. For example, in the quotient map $\pi\colon \mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$, the image of $\{0,1\}$ is a subgroup, but $\{0,1\}$ is not a subgroup of $\mathbb{Z}$.