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Given any function $f: X \to Y$ (and consider in general the set of functions with this domain and codomain), there is an induced function $f': \mathcal{P}(X) \to \mathcal{P}(Y)$ such that $f': A \mapsto f(A)$ etc. My question is about what structure is preserved between the two. Do we have to restrict the set of functions to those which are bijective in order to achieve isomorphism (and what does the isomorphism preserve? Inverses, corresponding to the preimage? What else? Is this an important relationship?

For example, do I always have $(g \circ f)' = g' \circ f'$? Do I need to restrict to bijective functions for this and/or anything else?

EE18
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    It is not clear what kind of "isomorphism" you want (isomorphism in which category? between which objects?). Only the final question is clear, and you should edit your post to include your attempts to answer it. – Anne Bauval Sep 12 '23 at 22:05
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    In general, the induced map $f^{-1}\colon \mathcal P(Y)\to \mathcal P(X)$ is better behaved. It will preserve unions, intersections, and complements. The forward map just preserves unions. (Though in the case of bijections the forward map also preserves all three of these, as it is inverse of the inverse). – M W Sep 12 '23 at 22:14
  • @AnneBauval I am working some exercises which just ask me about specific cases of operations on sets (MW mentions a few). I can include my proofs but I guess I was just looking for a "30000-foot" view of what's going on. – EE18 Sep 12 '23 at 22:25
  • It would good to include, if not your proofs themselves, then at least what you've proved so far (and also things that you haven't proved but suspect.), so that answerers will know what you've already figured out. – Toby Bartels Sep 14 '23 at 15:42
  • Very fair in general, but here I am not asking a question about an exercise, but more so a conceptual question/a question which occurred to me while reading the main body of text. @TobyBartels – EE18 Sep 14 '23 at 17:19
  • I didn't think that it was an exercise. I said that because 2 people have voted to close the question for a lack of details or focus, and it would really help to know if you have conjectures, to confirm or refute them. (But if you're happy with Qiaochu's answer, then maybe it doesn't matter.) – Toby Bartels Sep 15 '23 at 18:24
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    I see, thank you for the info! I didn't even know I could see close votes until now... @TobyBartels – EE18 Sep 15 '23 at 19:17

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It's not clear what you mean by either "structure" or "isomorphism" but here are some comments. I will write this induced map as $f_{\ast}$. It is called the direct image. The assignment $f \mapsto f_{\ast}$, together with the assignment $X \mapsto P(X)$, is a functor, meaning it satisfies $(f \circ g)_{\ast} = f_{\ast} \circ g_{\ast}$, with no further hypotheses.

$f_{\ast} : P(X) \to P(Y)$ has the property that it preserves arbitrary unions, meaning that

$$f_{\ast} \left( \bigcup_i X_i \right) = \bigcup_i f_{\ast}(X_i).$$

Conversely any function $P(X) \to P(Y)$ preserving arbitrary unions has the form $f_{\ast}$ but where $f$ is in general a relation rather than a function, which can be thought of as a function $X \to P(Y)$ which outputs subsets of $Y$ rather than elements.

One way to think about what $P(X)$ and $f_{\ast}$ are doing is that $P(X)$ is the free suplattice on $P(X)$ (a structure that formalizes the behavior of arbitrary unions) and $f_{\ast}$ is the map induced by this universal property.

Qiaochu Yuan
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  • Much of this is well above my head (it sounds like a lot of category theory) but it was interesting to learn about what little I understood -- thank you very much! It sounds like it is a useful construction. – EE18 Sep 13 '23 at 00:00