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I am having problems with establishing the following basic result. Actually, I found a previous post that is close in nature (it is about inverse image), but I was interested in this specific one, with the following notation, because it is the one I found in the book I am self-studying and I guess that most of my problems are actually related with the notation itself. Morevoer I would like to take it as an opportunity to find out how to write proof with equalities, because I tend in those cases to write them in a cumbersome way, with necessary and sufficient conditions.
[I assume that this is partly due to the influence of "How to prove it: a structured approach", a book that I loved, but that left me the tendency to be quite mechanical in proving any sort of result].

Theorem:
Let $X$ and $Y$ be nonempty sets and $f \in Y^X$. Prove that, for any (nonempty) classes $\mathcal{A} \subseteq 2^X$, we have $ f(\cup \mathcal{A}) = \cup \{f(A):A \in \mathcal{A} \} $

Here, it's how I approach the problem.

Tentative Proof:
First of all, by definition of (direct) image of a function, we have $$f(\cup \mathcal{A}) := \{f(x):x\in \cup \mathcal{A} \}. \hspace{1cm} (*)$$ Let $X$, $Y$ be arbitrary (nonempty) sets. Let $f$ be an arbitrary function that maps from $X$ to $Y$ and let $\mathcal{A}$ be a family of sets, subset of the powerset of $X$. By $(*)$, the result we have to prove becomes $$ \{f(x):x \in \cup \mathcal{A} \} = \cup \{ f(A):A \in \mathcal{A}\} \hspace{1cm} (1)$$ In order to prove it, rephrase $(1)$ as $$ \forall y ( y \in \{f(x):x \in \cup \mathcal{A} \} \leftrightarrow y \in \cup \{ f(A):A \in \mathcal{A}\}). \hspace{1cm} (2)$$ We start by proving the necessary condition. Let $y$ be an arbitrary element and assume that $y$ is a member of $\{f(x):x \in \cup \mathcal{A} \}$. This means that $\exists x(x \in \cup \mathcal{A} \land y=x)$. At the same time, we have to prove that $\exists A(A \in \mathcal{A} \land y\in f(A))$.

And here I got stuck...

I really don't see how from my premises I can get the desired result. Indeed, what $A$ should be?
I assume there is a problem with the way in which I translate the problem in logical terms.

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    $\begingroup$ Couldn't you just show both containments? That is usually a nice way to prove two sets are equal. For example, if $y$ is in the image of the union, it must be in the image of some $A$, so it is in the union of the images. The other containment is shown similarly. $\endgroup$
    – Jared
    May 20 '13 at 18:19
  • $\begingroup$ By $2^X$ do you mean the powerset of $X$, denoted $\mathcal{P}(X)$? They are naturally isomorphic but not equal. $\endgroup$ May 20 '13 at 19:07
  • $\begingroup$ Yes, $2^X$ I mean the powerset of $X$. $\endgroup$
    – Kolmin
    May 20 '13 at 19:27
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You’re confusing yourself by using more notation than is necessary. Suppose that $y\in f\left[\bigcup\mathscr{A}\right]$; then by definition there is an $x\in\bigcup\mathscr{A}$ such that $y=f(x)$. By the definition of union we know that $x\in A_x$ for some $A_x\in\mathscr{A}$, so $y=f(x)\in f[A_x]\subseteq\bigcup\left\{f[A]:A\in\mathscr{A}\right\}$. Since $y$ was arbitrary, this shows that

$$f\left[\bigcup\mathscr{A}\right]\subseteq\bigcup\left\{f[A]:A\in\mathscr{A}\right\}\;.$$

That’s all you have to say, and it’s far more readable than something cluttered with quantifiers and other formal logical notation.

The argument to show the reverse inclusion is essentially just running the same steps in reverse.

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  • $\begingroup$ The use of some sort of baroque notation is a bad habit I took from the book I mentioned. Usually when I write down proofs I try to avoid it, but when I am still in the - so to speak - working process, I cannot get rid of it, because it means that probably I still don't see the inner logic of the problem I am trying to solve, and I still need something really pedantic to move around and get the point. Anyway, thanks a lot for all those feedbacks. $\endgroup$
    – Kolmin
    May 20 '13 at 18:31
  • $\begingroup$ @Kolmin: You’re welcome. $\endgroup$ May 20 '13 at 18:33
  • $\begingroup$ Actually, in the end I notice that there was a flaw beyond the notation. Indeed I wrote $\exists x(x \in \cup \mathcal{A} \land y=x)$, but this is wrong. So, that's why I could not make it. $\endgroup$
    – Kolmin
    May 20 '13 at 18:37
  • $\begingroup$ @Kolmin: Right, you left out the $f$. I should have caught that, but I confess that I was concentrating more on the issue of clarity than on the actual details of the argument. $\endgroup$ May 20 '13 at 18:39
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The following are equivalent: $$y\in f\left(\bigcup\mathcal{A}\right)\\\exists x\in\bigcup\mathcal{A}:y=f(x)\\\exists A\in\mathcal A:\exists x\in A:y=f(x)\\\exists A\in\mathcal{A}:y\in f(A)\\y\in\bigcup\{f(A):A\in\mathcal{A}\}$$ You should be able to justify each step by definition of union or image.

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$$\frac{f(A)\subseteq B}{A\subseteq f^{-1}(B)}$$

Hence direct image is left-adjoint to inverse image, hence it preserves colimits, thus in particular it preserves unions.

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