If $f:X→X$ is a bijection then is $g(x):=\begin{cases}f(x),\,\text{if }x∈Y\\x,\,\text{otherwise}\end{cases}$ with $Y∈\mathcal P(X)$ a bijection too?

Question

Let be $f$ a bijection from a set $X$ to $X$ so that for any $Y$ in $\mathcal P(X)$ let's we put $$ g(x):=\begin{cases} f(x),&\text{if }x\in Y\\ x,&\text{otherwise} \end{cases} $$ with $x$ in $X$. So I observe that if $Y$ if finite and such that the inclusion $$ \tag{1}\label{1}f[Y]\subseteq Y $$ holds then $g$ is a bijection: indeed, if the last inclusion holds then (is this true?) by finitude of $Y$ we conclude that $f[Y]$ is exactly $Y$ and so if $x_1$ is in $X\setminus Y$ and $x_2$ in $Y$ then the same is for $g(x_1)$ and $g(x_2)$ so that they are different and this proves that $g$ is injective since if $x_1$ and $x_2$ are in $Y$ or in $X\setminus Y$ then trivially $g(x_1)$ and $g(x_2)$ are different when $x_1$ and $x_2$ are it; moreover if $f[Y]$ is $Y$ then for any $y$ in $Y$ there exists $x$ in $Y$ such that $y$ is $f(x)$ so that $g$ is even surjective since it is trivially surjective onto $X\setminus Y$. So first of all I ask if actually $g$ is a bjection when \eqref{1} holds and when $Y$ is finite and so I then ask if otherwise if $g$ is or not a bjection or rather, more explicitly, I ask (giving eventually a counterexample) if $g$ is a bijection even if \eqref{1} does not hold or if \eqref{1} holds but $Y$ is not finite. So could someone help me, please?

Answer 1

How about $X = \{1,2\}$ and $f(1) = 2, f(2) =1$ with $Y = \{1\}$. Then, $g(x) = 2$ for all $x$ hence $g$ is not a bijection.

If we include the extra condition (1) and assume that $Y$ is finite then we know $g$ is a bijection. However, let's not assume $Y$ is finite. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x) = x+1$ and take $Y = [0,\infty)$ then $f(Y) \subset Y$ and $g(x)$ is not a bijection because it is not surjective. For example, $1/2$ is not in the image of $g$.