# Continuous image of a Polish space to another has the Baire property

Here's a theorem in the course notes of a course on Polish groups:

Let $$X$$ and $$Y$$ be Polish spaces and $$f:X\to Y$$ continuous.

$$f(X)$$ has the Baire property.

In the course note, it's written that this follows from the existence of $$U(A)$$ (which I define bellow). The proof should be easy otherwise I would have written it. Unfortunately, while reviewing the notes, I forgot the proof and I'm unable to prove the theorem above.

Here's the definition of $$U(A)$$:

Let $$X$$ be a Polish space and $$A\subseteq X$$.

$$U(A):=\bigcup\{O\mid A\text{ is comeager in O, i.e, O\backslash A is meager}\}.$$

We have the following facts:

1. $$U(A)$$ is open.

2. $$A$$ is comeager in $$U(A)$$.

3. $$A$$ has the Baire property $$\iff$$ $$A\backslash U(A)$$ is meager.

Attempt

So I tried to show that $$f(X)\backslash U(f(X))$$ is meager.

If $$f(X)$$ is meager, we're done. Otherwise $$U(f(X))\neq\emptyset$$ and $$U(f(X))\backslash f(X)$$ is meager. I coudln't find a way to use the continuity of $$f$$ to conclude. I also tried to prove the theorem by contradiction using $$f(X)\backslash U(f(X))=\bigcap\{f(X)\backslash O\mid f(X)\text{ is comeager in O}\}$$

Let $$X$$ and $$Y$$ be Polish spaces and $$f:X\to Y$$ continuous. $$f(X)$$ has the Baire property.