Exist a homeomorphic copy of $\mathcal{C}$ contained in $f(X)$. Where $\mathcal{C}=2^\mathbb{N}$.

Question

If $X$ be a nonempty perfect Polish space, $Y$ a second countable space, and $f \colon X\to Y$ be injective and Baire measurable.

Question: Exist a homeomorphic copy of $\mathcal{C}$ contained in $f(X)$. Where $\mathcal{C}=2^\mathbb{N}$.

Any help I do not want to answer this question

Thanks.

Answer 1

Alexander S. Kechris. Classical Descriptive Set Theory, – Springer, 1995, page 52:

(8.38) Theorem. Let $X$, $Y$ be topological spaces and $f\colon X\to Y$ be Baire measurable. If $Y$ is second countable, there is a set $G\subseteq X$ that is a countable intersection of dense open sets such that $f|G$ is continuous. In particular, if $X$ is Baire, $f$ is continuous on a dense $G_\delta$-set.

Proof. Let $\{U_n\}$ be a basis for $Y$. Then $f^{-1}(U_n)$ has the BP in $X$, so let $V_n$ be open in $X$ and let $F_n$ be a countable union of closed nowhere dense sets with $f^{-1}(U_n)\triangle V_n\subseteq F_n$. Then $G_n=X\setminus F_n$ is a countable intersection of dense open sets and so is $G=\bigcap_n G_n$.
Since $f^{-1}(U_n)\cap G=V_n \cap G$, $f|G$ is continuous. $\square$

(8.39) Exercise. Let $X$ be a nonempty perfect Polish space, $Y$ a second countable space, and $f\colon X\to Y$ be injective and Baire measurable. Then there is a homeomorphic copy of $\mathcal C$ contained in $f(X)$.