-1
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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?

  • 1
    Don't you have a text or some notes or something where you could look up the definition of preimage?2012-11-04
  • 0
    http://en.wikipedia.org/wiki/Image_(mathematics\)2012-11-04
  • 0
    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