$f:X\rightarrow Y$. Set $B = Y - V$. Then, why is $f^{-1}(B) = f^{-1}(Y)-f^{-1}(V)$?
This statement is given in Topology (2e) by Munkres.
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2This has nothing to do with the topologies on $X$ and $Y$: If $X$ and $Y$ are sets and $f \colon X \to Y$ is a function, then $f^{-1}(Y \setminus A) = X \setminus f^{-1}(A)$ for every subset $A \subseteq Y$. – Jendrik Stelzner Feb 16 '16 at 18:57
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1This seems to be a duplicate of this question. You can find some related results here. – Martin Sleziak Feb 16 '16 at 22:44
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$x\in f^{-1}(B)$ i.e $f(x)\in Y-V$ i.e $f(x)\in Y$, $f(x)$ not $ V$ i.e $x\in f^{-1}(Y), x$ not in $f^{-1}(V)$ i.e $x\in f^{-1}(Y)-f^{-1}(V)$.
Tsemo Aristide
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