# Definition of surjectivity on topological spaces.

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 😃

• 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 – Cardioid_Ass_22 Mar 22 at 0:43
• @Cardioid_Ass_22 that clears everything up. Thank you :) – Patrick Mar 22 at 0:50

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!