Compact-test exponential law and products of quotient maps

For a compact Hausdorff test $(v,w):K\to Y{\times }_{k}C(Y,Z)$, consider $a\mapsto w(a)(v(a))$. If its value at $a$ lies in open $O$, regularity gives a closed neighbourhood $L$ of $a$ with $w(a)(v(L))\subseteq O$. The set $N=w^{-1}W(v{\mathrm{\mid }}_L,L,O)$ is open and contains $a$. For $b\in N\cap \mathrm{int}L$, one has $w(b)(v(b))\in O$. Thus every test composite of evaluation is continuous, so evaluation $Y{\times }_{k}C(Y,Z)\to Z$ is continuous.

Given continuous $g:X{\times }_{k}Y\to Z$, each function $f(x):y\mapsto g(x,y)$ is continuous by the slice map. For tests $v:L\to X$, $u:K\to Y$, the map $(b,a)\mapsto g(v(b),u(a))$ is continuous on the compact Hausdorff product. The tube lemma says that the set of $b$ for which all its values on $\{b\}\times K$ lie in $O$ is open. This is $(fv{)}^{-1}W(u,K,O)$. Testing on $L$ proves continuity $X\to C_0(Y,Z)$, and the CG-source property lifts it to $C(Y,Z)$.

Conversely, a continuous $f:X\to C(Y,Z)$ uncurries continuously by composing its product with the evaluation of step 1.1. The two constructions are inverse since both give the same value $f(x)(y)$ at every pair. Precomposition and postcomposition preserve this equality, proving naturality.

The map $(h,x,y)\mapsto h(x)(y)$ is a composite of two continuous evaluations. Twice applying the map correspondence of steps 1.2–2.1 makes the induced map $C(X,C(Y,Z))\to C(X{\times }_{k}Y,Z)$ continuous. Conversely, start with evaluation $(h,x,y)\mapsto h(x,y)$ and curry successively in $y$ and $x$. This gives the inverse continuous map. Product reassociations are homeomorphisms by F1, so the displayed bijection is a homeomorphism.

Let $P$ be the ordinary quotient of $X{\times }_{k}Y$ by the stated relation and $p$ its quotient map. It is CG. The map $q\times \mathrm{i}\mathrm{d}$ descends to a continuous bijection $a:P\to Q{\times }_{k}Y$. The transpose $x\mapsto (y\mapsto p(x,y))$ is constant on each q-fibre and hence descends continuously to $Q\to C(Y,P)$. Uncurrying gives $b:Q{\times }_{k}Y\to P$ with $b(q(x),y)=p(x,y)$. Surjectivity of q and p gives $ab=\mathrm{i}\mathrm{d}$ and $ba=\mathrm{i}\mathrm{d}$. Thus $q\times \mathrm{i}\mathrm{d}$ is quotient.

For quotient maps $q:X\to Q$ and $r:Y\to R$, factor $q\times r=({\mathrm{i}\mathrm{d}}_Q\times r)(q\times {\mathrm{i}\mathrm{d}}_Y)$. Each factor is quotient by step 3.2 and symmetry, and their composite is quotient. Empty factors give empty products and the same inverse identities; no representative of a quotient fibre has been selected.