I was trying to understand why
A regular ordinal is always an initial ordinal.
and in the course of this, came to the following hypothesis.
For any ordinals $\alpha$ and $\beta$ such that $$\omega_{\alpha} \le \beta < \omega_{\alpha + 1}$$ where $\omega_{\alpha}$ is the $\alpha$-th infinite initial ordinal and $\beta\,$ is a limit ordinal, we have $$\mathrm {cf} (\beta) = \mathrm {cf} (\omega_{\alpha}) = \begin{cases} \mathrm {cf} (\alpha), & \text{if $\alpha$ is a limit ordinal} \\ \omega_{\alpha}, & \text{otherwise} \end{cases}$$ Then we could simply use transfinite induction to prove the above fact since $\alpha < \omega_{\alpha} \le \beta$. Is the hypothesis true? If it isn't, how to prove the initial statement?
It is easy to see this is not the case. Since there are $\omega_{\alpha+1}$ many such $\beta$, it follows that for any $\mu<\alpha$, we can get $\mu$ many ordinals and take their supremum to construct an ordinal with cofinality less than $\operatorname{cf}(\omega_\alpha)$.
For example, we have
$$\omega_1\le\omega_1+\omega<\omega_2$$
but
$$\operatorname{cof}(\omega_1+\omega)=\omega$$
The gist of why regular ordinals are initial ordinals is that if it bijected a smaller ordinal, you could construct a cofinal sequence with length smaller than itself. More details here.