I would like to construct an explicit bijection from $\alpha^*$ to $\alpha$, where $\alpha$ is an infinite ordinal. I have been told a possible bijection would be:
\begin{equation} f(\beta) = \begin{cases} 0 & \text{if } \beta = \alpha \\ \beta^* & \text{if } \beta \text{ is finite} \\ \beta & \text{otherwise} \end{cases} \end{equation}
I don't understand how this is constructed, or why it is a bijection.
This is essentially a Hilbert's hotel argument.
The ordinal $\alpha^*$ is obtained by adding a new element to $\alpha$ (namely $\alpha$ itself, i.e. $\alpha^* = \alpha \cup \{ \alpha \}$). The idea is to define a function $f : \alpha^* \to \alpha$ that is as close to the identity function as possible.
We can't just define $f(\beta)=\beta$ for all $\beta \in \alpha^*$ since then $f(\alpha)$ would be undefined. So we define $f(\alpha)=0$. But then $f(0)$ can't be $0$, since then $f$ would not be injective, so we take $f(0)=1$. But then $f(1)$ can't be $1$ since $f$ would not be injective, so we take $f(1)=2$... and so on. But this 'and so on' stops at $\omega$, since the map $n \mapsto n+1$ gives a bijection $\omega \to \omega \setminus \{ 0 \}$.
So if we take $f(\alpha)=0$, $f(n)=n+1$ for all $n \in \omega$, and $f(\beta) = \beta$ for all other $\beta$, then we obtain a bijection $\alpha^* \to \alpha$.