Weak Hausdorff diagonals and closed quotients

If $X$ is WH, its points are closed by singleton tests. For $u:K\to X$ compact Hausdorff, $L=u(K)$ is closed and compact. The surjection $u:K\to L$ is closed: any closed subset of $K$ is compact Hausdorff and its image is closed by WH. For distinct $a,b\in L$, the disjoint closed fibres can be separated by disjoint open $U,V\subseteq K$. The sets $L\setminus u(K\setminus U)$ and $L\setminus u(K\setminus V)$ are disjoint open neighbourhoods of $a,b$. Therefore $L$ is Hausdorff.

Conversely suppose the diagonal is closed. For tests $u:K\to X$ and $v:L\to X$, the set $E=\{(a,b):u(a)=v(b)\}$ is closed in the compact Hausdorff $K\times L$, hence compact. Its projection to $L$ is compact and closed and is exactly $v^{-1}(u(K))$. Thus $u(K)$ is k-closed in $X$ and therefore closed. This proves WH.

A test into a subspace composed with its inclusion has image closed in the ambient WH space, hence closed in the subspace. Kification preserves tests and has finer topology, so it preserves WH. Step 1.1 also shows that each test-image mapping subbasic open is a compact-Hausdorff-subspace subbasic open; the converse uses the inclusion as test.

If CG $X$ is WH, test its diagonal by $(v,w):K\to X{\times }_{k}X$. At $a$ with $v(a)\mathrm{\ne }w(a)$, regularity provides a closed neighbourhood $L$ of $a$ inside $v^{-1}(X\setminus \{w(a)\})$. Then $v(L)$ is closed by WH, and $\mathrm{int}L\cap w^{-1}(X\setminus v(L))$ is a neighbourhood of $a$ on which $v\mathrm{\ne }w$. Each test equality set is closed; compact generation makes the diagonal closed.

The diagonal of a finite k-product is the intersection of the inverse images of the factor diagonals. Hence it is closed and the product is WH. A compact test into a finite coproduct has compact Hausdorff clopen inverse-image pieces; their images are closed in their summands, so their finite union is closed in the coproduct. Closed subspaces are CG by F2 and WH by step 2.1. These prove the asserted finite closure properties, including the empty product and coproduct.

For $q:X\to Q$, the quotient $Q$ is CG and $q\times q$ is quotient. Therefore its diagonal is closed exactly when its inverse image, the stated fibre equivalence relation, is closed. Steps 2.2–1.2 identify this condition with WH of $Q$.

Each point evaluation $e_x:C(X,Y)\to Y$ is continuous since inverse images of opens are singleton-test subbasic opens. The diagonal of $C(X,Y)$ is ${\bigcap }_{x\in X}(e_x\times e_x{)}^{-1}{\mathrm{\Delta }}_Y$, hence closed. This mapping space is CG by definition, so is WH. Requiring a basepoint or either interval endpoint to map to the closed point $y_0$ cuts out a closed subspace. Such subspaces are CGWH by step 3.1. The empty intersection when $X$ is empty gives the one-point mapping space.