Inverse bijections between subgroups of $G$ containing the kernel and subgroups of $H$ contained in the image. Claim: Let $\varphi:G \rightarrow H$ be a group homomorphism. Then the maps $A \mapsto \varphi(A)$ and $B \mapsto \varphi^{-1}(B)$ (where here we mean preimage) specify inverse bijections between the sets $\{A \leq G \mid \ker(\varphi) \subseteq A\}$ and $\{B \leq H \mid B \subseteq \text{Im}(\varphi)\}$.
Proof: Let $A \leq G$ contain the kernel, then it's evident that $A \subseteq \varphi^{-1}(\varphi(A))$. Let $g \in \varphi^{-1}(\varphi(A))$, then this means that $\varphi(g)= \varphi(a)$ for some $a \in A$. Then because $A$ is a subgroup $\varphi(ga^{-1}) = \varphi(g)\varphi(a^{-1}) = 1_H$ and so $ga^{-1} \in \ker(\varphi) \subseteq A$, but then because $A \leq G$ we have $(ga^{-1})a = g \in A$ and so the other inclusion holds.
The argument above isn't difficult, but I can't understand why we need to do this in the first place? If we're considering just images and preimages, doesn't it follow immediately from the definition that $(\varphi \circ \varphi^{-1})(A) = A =(\varphi^{-1} \circ \varphi)(A)$? What about the statement necessitates we prove it via equality of sets? Thanks in advance for the clarification.
 A: It isn't true that $(\varphi^{-1} \circ \varphi)(A)=A$.
Suppose $G = \mathbb{Z},\ A = \{0\},\ H = \{0\}$ and $\varphi(n) = 0\quad \forall n \in \mathbb{Z}$
then $(\varphi^{-1} \circ \varphi)(A) = G \neq A$.
A: *

*You only did half of the work: you didn't prove that forall subgroup $B$ in your second set, $\varphi(\varphi^{-1}(B))=B$.

*You did that first half well and (with the same argument) you could have proven that more generally $\forall X\subset G\quad\varphi^{-1}(\varphi(X))=(\ker\varphi)X$. Similarly, for the second "half of the work", you could easily prove more generally (and here, without using that $\varphi$ is a morphism) that $\forall Y\subset H\quad\varphi(\varphi^{-1}(Y))=Y\cap\operatorname{im}\varphi$.

*Using the two properties $\varphi^{-1}(\varphi(X))=(\ker\varphi)X$ and $\varphi(\varphi^{-1}(Y))=Y\cap\operatorname{im}\varphi$, it is easy to construct, for most morphisms $\varphi:G\to H$, a subgroup $A$ of $G$ and a subgroup $B$ of $H$ such that $A\subsetneq\varphi^{-1}(\varphi(A))$ and $\varphi(\varphi^{-1}(H))\subsetneq H$. Exercise: for which morphisms do such counterexamples exist?

