functions - Prove the existence of a bijection

Let $A$ and $B$ be sets, and suppose $A$ is infinite.
Let $B$ be a countably infinite subset of $A$.

Show that if $f: \mathbb{N} \to B$ and $g: B \to \mathbb{N}$ are bijections, then
$$
h: A \to A- \{ f(1) \}, \, a \mapsto
\begin{cases}
a, & \text{if $a \notin B$} \\
f(1+g(a)), & \text{if $a \in B$ }

\end{cases},
$$
is also a bijection.

I am not really sure how to do this one

Answer

To show injectivity, assume that $h(x) = h(y)$ for some $x,y \in A$. We want to show that $x = y$. There are a few cases:

Case 1 ($x,y \in B$): Then by definition $h(x) = x$ and $h(y) = y$. So we have $x = y$.

Case 2 ($x,y \notin B$): Then we have that $f(1+g(x)) = f(1+g(y))$. Since both $f,g$ are invertible, we have that $x = y$.

Case 3 ($x \in B$, $y \notin B$): Then $h(x) \in B$ since $f(1+g(x)) \in B$, and $h(y) = y \notin B$. Thus $h(x) \neq h(y)$, so this case is impossible.

Case 4 ($x \notin B$, $y \in B$): As in case 3, this case cannot happen.

So we have that $h(x) = h(y) \Rightarrow x = y$. Thus $h$ is injective.

Now we want to show surjectivity. Let $b \in A - f(1)$. We want to show that there exists an $a \in A$ such that $h(a) = b$. There are a couple cases:

Case 1 ($b \notin B$): Set $a = b$. Then $h(a) = h(b) = b$.

Case 2 ($b \in B$): Then $b = f(n)$ for some $n \in \mathbb{N}$, since $f$ is bijective. Further, $n > 1$ since $b \in A - f(1)$. We have $g(a) = n-1$ for some $a \in B$, since $g$ bijective. Then $h(a) = f(n) = b$.

Thus $h$ is surjective. Since $h$ is both injective and surjective, it is bijective.