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If a map $f$ between topological spaces, $X$ and $Y$, is surjective, does that mean that every open set in the topological space $Y$ has an open preimage, or merely that the set $Y$ has an open preimage? I guess my question is: is surjectivity mean hitting all the the elements of the set $Y$, or all the open sets on the topological space $Y$? Does it depend on the context? Thanks 😃

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    $\begingroup$ Technically both are true- surjectivity means that each element in $Y$ is hit at least once so all open sets are also 'hit' (by subsets of $X$) but the preimage of each open set in $Y$ will always be open in $X$ iff $f$ is continuous $\endgroup$ – Cardioid_Ass_22 Mar 22 at 0:43
  • $\begingroup$ @Cardioid_Ass_22 that clears everything up. Thank you :) $\endgroup$ – Patrick Mar 22 at 0:50
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Unless I'm missing some non-standard definition, surjectivity is not a topological property. Whether a map between two topological spaces $X$ and $Y$ is surjective does not depend one jot on the topology of $X$ or the topology of $Y$.

To say $f : X \to Y$ is surjective is to say that, for every $y \in Y$, there exists an $x \in X$ such that $f(x) = y$. It may not be the case that open subsets of $Y$ have open preimages (i.e. $f$ need not be continuous). It will, however, always be the case that $f^{-1}(Y) = X$, which is open, as every element of $X$ maps into $Y$, by definition of a function. This is true even when $f$ is not surjective!

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