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Simple question: Suppose $f:X\to Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $y\in Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?

In general, for $f:X\to Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?

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  • $\begingroup$ $f^{-1}(y)$ always exists. It is the set of all $x \in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = \emptyset$ if $y $ is not in the image f $X$. $\endgroup$
    – Paul Frost
    Jan 14, 2019 at 14:59
  • $\begingroup$ @PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}(\{y,z\})=\{a,b,c\}$? $\endgroup$
    – metalder9
    Jan 14, 2019 at 17:56
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    $\begingroup$ Correct! For any $B \subset Y$ the defnition is $f^{-1}(B) = \{ x \in X \mid f(x) \in B \}$. $\endgroup$
    – Paul Frost
    Jan 14, 2019 at 18:00

1 Answer 1

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If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.

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