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When it comes to talking about maths, it's nice to be able to express mathematical ideas concisely and eloquently. When speaking of the fact that, for any map of sets $f:X \rightarrow Y$, we have that $ \bigcup f^{-1}(U_i) = f^{-1}\left(\bigcup U_i\right)$, I want to say that "taking unions of sets in the codomain commutes with taking the pre-image", or something like that. I've heard more experienced mathematicians say stuff along those lines before, and was wondering whether I can say that here.

I really don't know if that's correct language. Is it?

Matt
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  • It cannot commute because the other union is in the domain. Perhaps losely speaking: the union of the pre-images is the pre-image of the union". – max_zorn Jan 08 '18 at 08:29
  • In ancient India or Greece, mathematicians exclusively expressed their thoughts like this, formulas weren't invented, yet. –  Jan 08 '18 at 08:34
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    “Preimages commute with unions” would be fine. Another way to say it is that preimages preserves unions. – Qiaochu Yuan Jan 08 '18 at 08:35

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To teach this topic verbally, I often state a kind of a maxim:

Functions commute with unions and inverses of functions commute with Boolean operations.

Boolean operations include union, intersection and complement, and hence the second part of the maxim means, in mathematical notation, $$ \bigcup_{i\in I} f^{-1}(U_i) = f^{-1}\Bigl(\bigcup_{i\in I} U_i\Bigr)\qquad \bigcap_{i\in I} f^{-1}(U_i) = f^{-1}\Bigl(\bigcap_{i\in I} U_i\Bigr)\qquad f^{-1}(U^c) = \bigl(f^{-1}(U)\bigr)^c $$

J.-E. Pin
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