# Transferability of space properties via continuous functions

Let $f:(X, \tau_x) \to (Y,\tau_y)$ be a continuous and onto function. I need to show that if $X$ is separable (Lindelöf), then $Y$ is respectively separable (Lindelöf).

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Klara, have you written down the definitions of separable(Lindelof) and continuous functions yet? That would be a good start. Also, it might help people find your question if you edit the title to mention either something about separable and/or the fact that it's not properties of "continuous functions" you want to transfer, but rather properties of "spaces" via a continuous function. – Mark S. Nov 13 '12 at 4:26
@ helopticor Thank you. – Klara Nov 13 '12 at 4:47
Continuity of $f$ doesn’t imply that $f[\operatorname{cl}D]=\operatorname{cl}f[D]$; it guarantees only that $f[\operatorname{cl}D]\subseteq\operatorname{cl}f[D]$. In this case you do get equality, but only because $\operatorname{cl}D=X$ and $f$ is onto. I’d have argued simply that if $U$ is any non-empty open set in $Y$, then $f^{-1}[U]$ is a non-empty open set in $X$, so $D\cap f^{-1}[U]\ne\varnothing$. Let $x\in D\cap f^{-1}[U]$; then $f(x)\in f[D]\cap U$, so $f[D]$ is dense in $Y$. – Brian M. Scott Nov 16 '12 at 20:59
@ Brian you are right, I just checked my book it sounded good to assume equality:), sorry.Thank you so much! – Klara Nov 16 '12 at 22:55

You need to assume in addition that $f$ maps $X$ onto $Y$.
(1) Let $D$ be a countable dense subset of $X$, and consider the set $f[D]$ in $Y$. It’s certainly countable; can you show that it’s dense?
(2) Let $\mathscr{U}$ be an open cover of $Y$, and let $\mathscr{V}=\{f^{-1}[U]:U\in\mathscr{U}\}$; use the continuity of $f$ to conclude that $\mathscr{V}$ is an open cover of $X$. Let $\mathscr{W}$ be a countable subcover of $\mathscr{V}$; can you see where to go from here to finish up?
@Klara: Or for each $V\in\mathscr{V}$ you could pick a $U_V\in\mathscr{U}$ such that $V=f^{-1}[U_V]$ and let $\mathscr{G}=\{U_V:V\in\mathscr{V}\}$, but since $f$ is onto, it comes to the same thing and does indeed give you the desired countable subcover. – Brian M. Scott Nov 16 '12 at 20:55