Compactly generated spaces as quotients of topological sums of compact Hausdorff spaces I have two questions about Proposition 7.9.2 and Corollary 7.9.3 in tom Dieck's "Algebraic Topology".
Here is the setting, taken from Tammo tom Dieck: Algebraic Topology, European Mathematical Society, 2008 (Section 7.9):

A compact Hausdorff space will be called a ch-space. For the purpose of the
following investigations we also call a ch-space a test space and a continuous map $f: C \to X$ of a test space $C$ a test map. [...]


A subset $A$ of a topological space $(X, \mathcal{T})$ is said to be k-closed (k-open), if for each test map $f: K \to X$ the pre-image $f^{-1}(A)$ is closed (open) in $K$. The k-open sets in $(X, \mathcal{T})$ form a topology $k\mathcal{T}$ on $X$. A closed (open) subset is also k-closed (k-open). Therefore $k\mathcal{T}$ is finer than $\mathcal{T}$ and the identity $\iota=\iota_X: kX \to X$ is continuous. We set $kX = k(X) = (X, k\mathcal{T})$. [...]


A topological space $X$ is called k-space, if the k-closed sets are closed, i.e., if $X = kX$. [...]


The next proposition explains the definition of a k-space. We call a topology $\mathcal{S}$ on $X$ ch-definable, if there exists a family $(f_j: K_j \to X \,|\, j \in J)$ of test maps such that: $A \subset X$ is $\mathcal{S}$-closed $\Leftrightarrow$ for each $j \in J$ the pre-image $f_j^{-1}(A)$ is closed in $K_j$. We can rephrase this condition: The canonical map $\langle f_j \rangle: \bigsqcup_j K_j \to (X,\mathcal{S})$ is a quotient map. A ch-definable topology is finer than $\mathcal{T}$. We define a partial ordering on the set of ch-definable topologies by $\mathcal{S}_1 \leq \mathcal{S}_2 \Leftrightarrow \mathcal{S}_1 \supset \mathcal{S}_2$.


(7.9.2) Proposition. The topology $k\mathcal{T}$ is the maximal ch-definable topology with respect to the partial ordering.


Proof. By Zorn’s Lemma there exists a maximal ch-definable topology $\mathcal{S}$. If this topology is different from $k\mathcal{T}$, then there exists an $\mathcal{S}$-open set $U$, which is not
k-open. Hence there exists a test map $t: K \to X$ such that $t^{-1}(U)$ is not open. If we adjoin this test map to the defining family of $\mathcal{S}$, we see that S is not maximal.


(7.9.3) Corollary. The k-spaces are the spaces which are quotients of a topological sum of ch-spaces.

My questions:
(1) Why do we need Zorn's Lemma? Isn't it true by definition of $k\mathcal{T}$ that this is the coarsest topology on $X$ which is ch-definable? Thus, in particular, it should be a maximal element with respect to the above partial ordering, or what am I missing?
(2) Why is every k-space (homeomorphic to) a quotient of a disjoint union of ch-spaces? Since $k\mathcal{T}$ is a ch-definable topology on $X$ (essentially by definition), there exists a quotient map of the form $\langle f_j \rangle: \bigsqcup_j K_j \to (X,k\mathcal{T})$, so if $X$ is a $k$-space, then $X=kX=(X,k\mathcal{T})$ is bijective to such a quotient. But why is there a homeomorphism? Sorry if this is obvious.
Any help is much appreciated. Many thanks in advance!
 A: Using the comments, I would like to answer my first question:
The class of all test maps to $X$ is not in general a set, so it is not clear from the definition of $k\mathcal{T}$ that this topology is ch-definable. Tom Dieck uses Zorn's Lemma in order to obtain a maximal ch-definable topology on $X$ with respect to the partial ordering he has defined, and then he shows that this topology is $k\mathcal{T}$. In particular, this implies that $k\mathcal{T}$ is in fact ch-definable.
And I think I have found an answer to my second question:
If $X=(X,\mathcal{T})$ is a k-space, that is, $\mathcal{T}=k\mathcal{T}$, then by the proposition, its topology is ch-definable, so there exists a quotient map $q: Z\to X$ for some topological sum $Z$ of ch-spaces. We define an equivalence relation $\sim$ on $Z$ by $z_1\sim z_2 \Leftrightarrow q(z_1)=q(z_2)$. Then $q$ induces a well-defined bijective continuous map $f:Z/\!\sim\, \to X$ satisfying $f\circ p = q$ where $p:Z\to Z/\!\sim$ is the canonical projection.
This map $f$ is a homeomorphism: Given an open subspace $U\subset Z/\!\sim$, we need to show that the image $f(U)$ is open in $X$, or equivalently, that $q^{-1}(f(U))$ is open in $Z$ (since $q:Z\to X$ is a quotient map). This is true because $q^{-1}(f(U)) = (f\circ p)^{-1}(f(U)) = p^{-1}(f^{-1}(f(U))) = p^{-1}(U)$ (by injectivity of $f$), which is open in $Z$ (by definition of the quotient topology on $Z/\!\sim$).
Hence, $X$ is homeomorphic to the quotient $Z/\!\sim$.
