Does Composition of compressing Collision Resistant Hash Functions $H^{*}=H_2(H_1(x))$ is Collision Resistant?
Assume that, $H_1 \text{ and } H_2$ are collision resistant and $H^*$ is not. We will show that one of the $H_1 \text{ and } H_2$ can not be a collision resistant.

Let $x_1 \neq x_2$ be inputs such that $$H^*(x_1) = H^*(x_2).$$ $$H_2(H_1(x_1)) = H_2(H_1(x_2)).$$i.e we have a collision pair.

Now, if $H_1$ is collision resistant then $$y_1 = H_1(x_1) \neq H_1(x_2) = y_2 .$$

Note : The equality holds only with negligible probability, then $H_2(H_1(x_1)) = H_2(H_1(x_2))$ is a collision with a negligible probability.

Now, $H_2(y_1) = H_2(y_2)$ since $H^*$ is not collision resistant. But we found a collision for $H_2$ given a collision for $H^*$. This is a contradiction since $H_2$ is collision resistant therefore this implies that $H^*$ is collision resistant, too.