general topology - Using Zorn's lemma for a proof about irreducible continuous functions.

Let $f:X\rightarrow Y$ onto and continuous between two Hausdorff compact spaces. We say that $f$ is irreducible iff for any closed proper subset $H\subset X$, $f(H)\neq Y$.

Now, consider an arbitrary, onto and continuous function $f:X\rightarrow Y$ where $X$ and $Y$ are Hausdorff compact spaces. I need to show that there exists a closed set $Z\subseteq X$ such that $f\upharpoonright Z$ is irreducible.

My attempt: Let $\Sigma=\{Z\subseteq X:Z\ \text{is closed}\ \wedge f\upharpoonright Z\ \text{is onto}\}$. It is clear that $(\Sigma,\supset)$ is a partial order for $\Sigma$ and $\Sigma\neq\emptyset$ because $X$ is one of its elements. Let $C\subseteq\Sigma$ a chain and as $X$ is a compact space, $\bigcap C\neq\emptyset$ (because $C$ has FIP)...

As you can see, I try to use Zorn's lemma in my proof. My problem is in the last part, I have many troubles when I try to show that $f\upharpoonright\bigcap C$ is onto, I don't see as I can get it (well, I think that $\bigcap C$ must be a lower bound for $C$ in $\Sigma$). Can you give me a hint?

Answer

If your chain is denoted $C_i, i \in I$ and we have that for all $i,j$ we either have $C_i \subseteq C_j$ or reversely. (This is what being a chain means). We want to show that $C = \bigcap_i C_i$ is the required upperbound in $\Sigma$ and for this we need that $f$ maps $C$ onto $Y$. For this consider $y \in Y$ and note that $D_i := C_i \cap f^{-1}[\{y\}]$ consists of closed subsets of $X$ (here we need that $Y$ is $T_1$ but that follows from Hausdorffness, and that $f$ is continuous) that are all non-empty (as every $C_i$ has at least one point mapping to $y$, as $f|_{C_i}$ is onto). We still have a chain, and hence the FIP holds and we have some $x \in \bigcap_i D_i \subseteq C$ by compactness. As $x \in D_i$ (for any $i$), in particular $f(x) = y$ and as $y \in Y$ was arbitrary, $f|_C$ is onto and the required Zorn chain-upperbound (as we have reverse inclusion as order).

So Zorn gives us a maximal element $M$ in $\Sigma$ and this is by definition as required: being in $\Sigma$ implies $f[M]=Y$ and if we have any closed $H \subsetneq M$, it has to obey $f[M] \neq Y$ or else it would be a member of $\Sigma$ and thus a properly "larger" element than the maximal $M$, which cannot be.