Are there two kinds of polynomial hierarchy collapses?
Suppose that $\Sigma_i^p = \Sigma_{i+1}^p$ (and so $\Pi_i^p = \Pi_{i+1}^p$). Then every statement of the form $(\exists y_1) \cdots (Q y_{i+1}) f(x,\vec{y})$ for polytime $f$ is equivalent to another statement $(\exists y_1) \cdots (Q y_i) g(x,\vec{y})$ for polytime $g$.

Now suppose you have a predicate in $\Sigma_{i+2}^p$, say $(\exists y) (\forall z_1) \cdots (Q z_{i+1}) f(x,y,\vec{z})$. The inner $i+1$ quantifiers are a $\Pi_{i+1}^p$ predicate, and so they are equivalent to a $\Pi_i^p$ predicate $(\forall z_1) \cdots (Q z_i) g(x,y,\vec{z})$. Adding back the outer quantifier $(\exists y)$, we find that the original predicate is equivalent to a $\Sigma_{i+1}^p$ predicate, and so $\Sigma_{i+2}^p = \Sigma_{i+1}^p = \Sigma_i^p$.

In the same way, we show that $\Sigma_i^p = \Sigma_{i+1}^p$ implies $\Sigma_i^p = \Sigma_{i+j}^p$ for all $j$, so that $\mathsf{PH} = \Sigma_i^p$.

Note that $\Pi_i^p \subseteq \Sigma_{i+1}^p$. Indeed, if $(\forall y_1) \cdots (Q y_i) f(x,\vec{y})$ is a $\Pi_i^p$ predicate, then the equivalent $(\exists w) (\forall y_1) \cdots (Q y_i) f(x,\vec{y})$ is a $\Sigma_{i+1}^p$ predicate.

Since $\Pi_i^p \subseteq \Sigma_{i+1}^p$, we see that $\Sigma_i^p = \Sigma_{i+1}^p$ implies that $\Pi_i^p \subseteq \Sigma_i^p$ and so $\Pi_i^p = \Sigma_i^p$.



Now for the other direction. Suppose that $\Pi_i^p = \Sigma_i^p$, and consider some $\Sigma_{i+1}^p$ predicate $(\exists y_1) (\forall y_2) \cdots (Q y_{i+1}) f(x,\vec{y})$. The inner $i$ quantifiers are a $\Pi_i^p$ predicate, and so they are equivalent to some $\Sigma_i^p$ predicate $(\exists z_2) \cdots (Q z_{i+1}) g(x,\vec{y})$. Adding back the outer quantifier $(\exists y_1)$ and folding $y_1,z_2$, we obtain a $\Sigma_i^p$ predicate equivalent to our original $\Sigma_{i+1}^p$ predicate.

Summarizing, if $\Pi_i^p = \Sigma_i^p$ then $\Sigma_i^p = \Sigma_{i+1}^p$, and so (as we have seen above) $\mathsf{PH} = \Sigma_i^p$.