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$f(S \cap T) \neq f(S) \cap f(T)$

but

$f^{-1}(S \cap T)$ = $f^{-1}(S) \cap f^{-1}(T) $

where $f^{-1}$ is a preimage

what is a preimage and what difference does it make?

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    to answer your title question, it would be a contradiction if the expression after the "but" were identical to the first expression, except for the equal sign. Here you have $f$ in the first expression, $f^{-1}$ in the second expression, so no, it's not a contradiction.2012-11-04

2 Answers 2

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If you have a function $f$ from some set $X$ to some set $Y$, and you have a subset $S$ of $Y$, then the preimage of $S$ is all those $x$-values in $X$ such that $f(x)$ is in $S$. Now I suggest you make up some examples of functions and see for yourself what $f(S\cap T)$ and $f^{-1}(S\cap T)$ and the rest look like.

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    I'm sorry, Gladstone, you're going to have to communicate in complete grammatical sentences if I'm to understand you (also if you're to understand mathematics).2012-11-04
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A pre-image is the set of all points that map to some point in the target set.

For example, if $f(x) = 1$ if $x$ is positive and even, and $f(x) = 0$ if $x$ is positive and odd, then the pre-image of $f(x)$ is set of all positive integers. Do you see why?

This is different from an inverse. An inverse, if it exists, will be such that $f^{-1}(f(x)) = x$. Unfortunately, we tend to re-use the notation $f^{-1}$. Remember that a function may not be invertible; however, every function has a pre-image.

Using this, can you solve the problem?