I am having trouble understanding a proposition about group homomorphisms. The book I am using is in portuguese, so I shall translate below the proposition:
Let $f: (G, *) \to (\mathcal{G}, \times)$ be a homomorphism between these two groups, and let $\mathcal{H}$ be a subgroup of $\mathcal{G}$. Then $f^{-1}(\mathcal{H})$ is a subgroup of G containing ker f and $f(f^{-1}(\mathcal{H})) = \mathcal{H}\cap\mathrm{im}\:{f}$.
While I do understand we never supposed f to be surjective, I am having trouble understanding why the intersection with the image was necessary in the last part of the statement. Isn't $\mathcal{H}$ already $\in\mathrm{im}\:{f}$, since the book made that statement about its inverse?
Attached is the original text in portuguese, in case I mistranslated something.
The issue is in your clause
which presumably means that you're reading "$f^{-1}(\mathcal{H})$" as implicitly requiring that $\mathcal{H}\subseteq im(f)$ (note: $\subseteq$, not $\in$). This is however incorrect.
For any function $f:A\rightarrow B$ and any set $C\subseteq B$ whatsoever, we can define the preimage $$f^{-1}(C):=\{x\in A: f(x)\in C\}.$$ Note that a given $x\in C$ might "pull back" to several elements of $A$, or to none at all; despite the notation, there's no assumption here that $f$ is "nice" when (co)restricted to $C$. In particular, note that we always have $f(f^{-1}(C))\subseteq C$ but we do not always have $f(f^{-1}(C))=C$. (For instance, maybe $C$ is "big" but disjoint from $im(f)$.)
On the plus side, this means your confusion has nothing to do with group theory, but rather just overloaded function notation!