proof of Baire space is universal for Polish spaces
Let $X$ be a nonempty Polish space^{}. We construct a continuous^{} onto map $f:\mathcal{N}\to X$, where $\mathcal{N}$ is Baire space^{}.
Let $d$ be a complete metric on $X$. We choose nonempty closed subsets $C({n}_{1},\mathrm{\dots},{n}_{k})\subseteq X$ for all integers $k\ge 0$ and ${n}_{1},\mathrm{\dots},{n}_{k}\in \mathbb{N}$ satisfying the following.

1.
$C()=X$.

2.
$C({n}_{1},\mathrm{\dots},{n}_{k})$ has diameter no more than ${2}^{k}$.

3.
For any $k\ge 0$ and ${n}_{1},\mathrm{\dots}{n}_{k}\in \mathbb{N}$ then
$$C({n}_{1},\mathrm{\dots},{n}_{k})=\bigcup _{m=1}^{\mathrm{\infty}}C({n}_{1},\mathrm{\dots},{n}_{k},m).$$ (1)
This can be done by induction^{}. Suppose that the set $S=C({n}_{1},\mathrm{\dots},{n}_{k})$ has already been chosen for some $k\ge 0$. As $X$ is separable^{}, $S$ can be covered by a sequence of closed sets ${S}_{1},{S}_{2},\mathrm{\dots}$. Replacing ${S}_{j}$ by ${S}_{j}\cap S$, we suppose that ${S}_{j}\subseteq S$. Then, remove any empty sets^{} from the sequence. If the resulting sequence is finite, then it can be extended to an infinite^{} sequence by repeating the last term. We can then set $C({n}_{1},\mathrm{\dots},{n}_{k},{n}_{k+1})={S}_{{n}_{k+1}}$.
We now define the function $f:\mathcal{N}\to X$. For any $n\in \mathbb{N}$ choose a sequence ${x}_{k}\in C({n}_{1},\mathrm{\dots},{n}_{k})$. Since this has diameter no more than ${2}^{k}$ it follows that $d({x}_{j},{x}_{k})\le {2}^{k}$ for $j\ge k$. So, the sequence is Cauchy (http://planetmath.org/CauchySequence) and has a limit $x$. As the sets $C({n}_{1},\mathrm{\dots},{n}_{k})$ are closed, they contain $x$ and,
$$\bigcap _{k=1}^{\mathrm{\infty}}C({n}_{1},\mathrm{\dots},{n}_{k})\ne \mathrm{\varnothing}.$$  (2) 
In fact, this has diameter zero, and must contain a single element, which we define to be $f(n)$.
This defines the function $f:\mathcal{N}\to X$. We show that it is continuous. If $m,n\in \mathcal{N}$ satisfy ${m}_{j}={n}_{j}$ for $j\le k$ then $f(m),f(n)$ are in $C({m}_{1},\mathrm{\dots},{m}_{k})$ which, having diameter no more than ${2}^{k}$, gives $d(f(m),f(n))\le {2}^{k}$. So, $f$ is indeed continuous.
Finally, choose any $x\in X$. Then $x\in C()$ and equation (1) allows us to choose ${n}_{1},{n}_{2},\mathrm{\dots}$ such that $x\in C({n}_{1},\mathrm{\dots},{n}_{k})$ for all $k\ge 0$. If $n=({n}_{1},{n}_{2},\mathrm{\dots})$ then $x$ and $f(n)$ are both in the set in equation (2) which, since it is a singleton, gives $f(n)=x$. Hence, $f$ is onto.
Title  proof of Baire space is universal for Polish spaces 

Canonical name  ProofOfBaireSpaceIsUniversalForPolishSpaces 
Date of creation  20130322 18:46:57 
Last modified on  20130322 18:46:57 
Owner  gel (22282) 
Last modified by  gel (22282) 
Numerical id  4 
Author  gel (22282) 
Entry type  Proof 
Classification  msc 54E50 