Let $X, Y$ be two metric spaces and $f$ be a continuous bijection (i.e. one-to-one map) from $X$ to $Y$. Let $E$ be a $G\_{\delta}$ subset of $X$. I want to know weather the image $f(E)$ is also a $G\_{\delta}$ subset of $Y$. Thanks!



On the image of a G_\delta set under a continuous bijection
Yes, if $X$ is $\sigma$-compact: $X\smallsetminus E$ is $F\_\sigma$, hence a countable union of compact sets. Since a continuous image of a compact set is compact (and therefore closed, in a Hausdorff space), $f(X\smallsetminus E)$ is also $F\_\sigma$, hence $f(E)$ is $G\_\delta$.


EDIT: While ljjpfx seems to be happy with this answer as his or her spaces are compact, Nate Eldredge asked in a comment whether there is a counterexample with both spaces Polish, which I think is an interesting question. I believe the following example should work. Consider the function
$$f\colon\omega^\omega\to[0,1),\quad f(a\_0,a\_1,a\_2,\dots)=0{.}\underbrace{1\cdots1}\_{a\_0}0\underbrace{1\cdots1}\_{a\_1}0\underbrace{1\cdots1}\_{a\_2}0\cdots,$$
where the right-hand side is meant to be the binary representation. (I will write $0{.}1^{a\_0}01^{a\_1}0\cdots$ to simplify the notation.) It is easy to see that $f$ is a continuous bijection, while $f^{-1}$ is not continuous. Let $U$ be the open set
$$\{a=(a\_n)\_{n\in\omega}\in\omega^\omega:\exists n\\,a\_n\text{ is even}\},$$
I claim that $f(U)$ is not $G\_\delta$. Assume for contradiction that $f(U)=\bigcap\_nG\_n$ for some open $G\_n\subseteq[0,1)$. Note that $f(U)$ is the union of intervals of the form $[a2^{-k},(a+1)2^{-k})$, where $a,k\in\omega$, and $a$ has the binary representation $1^{a\_0}01^{a\_1}0\cdots01^{a\_r}0$, where $a\_i$ are odd for $i< r$, $a\_r$ is even, and $k=\sum\_{i\le r}(a\_i+1)$. Since $G\_n\supseteq f(U)$ is open, for every $n,k$ there exists $l\_{n,k}$ such that
$$G\_n\supseteq(a2^{-k}-2^{-(k+l\_{n,k})},(a+1)2^{-k})$$
for every $a=1^{a\_0}01^{a\_1}0\cdots01^{a\_r}0$ as above. Note that if $a\_r>0$, the end-point of this interval has the form $0{.}1^{a\_0}01^{a\_1}0\cdots01^{a\_{r-1}}01^{a\_r-1}01^{l\_{n,k}+1}$. Choose $a=(a\_n)\in\omega^\omega$ so that every $a\_n$ is odd, and $a\_{n+1}>l\_{n,k\_n}$, where $k\_n=1+\sum\_{i\le n}(a\_i+1)$. Then $a\notin U$, but the description above implies that $f(a)\in\bigcap\_nG\_n$, a contradiction.


