 Solution

For every ultrafilter $\mathcal{H}$ on $X$, its image is an ultrafilter on the compact space $Y$, so it has a limit. That limit is unique because $Y$ is Hausdorff: distinct limits would have disjoint neighborhoods both belonging to the same proper filter. Therefore $\Lambda$ is well-defined. Since $f[X]$ is dense, every $b\in Y$ lies in $\overline{f[X]}$. Part (a), with $A=X$, supplies an ultrafilter mapping to $b$, proving surjectivity.

Suppose disjoint subsets have disjoint image closures. If $\mathcal{H}\ne \mathcal{K}$, choose $A\in \mathcal{H}$ not in $\mathcal{K}$; the ultrafilter criterion gives $X\setminus A\in \mathcal{K}$. Part (a) givesThe two closures are disjoint by assumption, so the limits differ. Thus $\Lambda$ is injective.

Conversely, suppose $\Lambda$ is injective and disjoint $A,B\subseteq X$ had a common point $b$ in their image closures. Part (a) would supply ultrafilters $\mathcal{H},\mathcal{K}$ containing $A,B$ respectively, both with image limit $b$. They must be distinct, since a proper filter cannot contain the disjoint sets $A$ and $B$ simultaneously. This contradicts injectivity. Hence