help checking proof of $f^{*}(f_{*}(\mathcal{D}(f)))=\mathcal{D}(f)$, where $\mathcal{D}(f)$ is the domain of a given function f

Let $f \colon \mathcal{D}(f) \subset A \to B$ be the mapping of sets $\mathcal{D}(f)$ and B. Define $f^{*}(B)$ as the pre-image of f under B, and $f_{*}(\mathcal{D}(f))$ as the image of f under $\mathcal{D}(f)$.

My progress is the following:

$\subseteq$: Let $x \in f^{*}(f_{*}(\mathcal{D}(f)))$. Therefore, $y = f(x) \in f_{*}(\mathcal{D}(f))$. And thus $x \in \mathcal{D}(f)$.

$\supseteq$: Let $x \in \mathcal{D}(f)$. Since x is in the domain, there exists an $f(x) \in f_{*}(\mathcal{D}(f))$. Let $\mathcal{J} = \mathcal{D}(f)$, and thus $f(x) \in f_{*}(\mathcal{J}) = \mathcal{H}$ Therefore $x \in f^{*}(\mathcal{H}) = f^{*}(f_{*}(\mathcal{D}(f)))$.

By double inclusion, the proof is complete.

Please give me any type of feedback you can think of, it is encouraged.

Solution 1:

For the $\subseteq$ direction: I think your reasoning does not quite make sense, or at least beats around the bush a bit. More directly, this direction holds because (using your notation) $f^*(S) \subset \mathcal{D}(f)$ for all subsets $S \subset B$, in turn because the definition of $f^*$ literally is just $$ f^*(S) := \{x \in \mathcal{D}(f) : f(x) \in S\}, $$ so is always a subset of $\mathcal{D}(f)$ by construction. So we just apply this to the special case where $S = f_*(\mathcal{D}(f))$.

For the $\supseteq$ direction: I think the wording "...there exists an $f(x) \in f_*(\mathcal{D}(f))$." is at least confusing, because there doesn't really "exist" such $f(x)$ as much as $f(x)$ is simply part of the data of the function $f$. So using the words e.g. "...we have..." would make more sense here than "...there exists an...". The rest seems to make sense to me. :)