The topological universal property

Proof. By the characteristic property of the quotient topology, a map out of $E_A$ is continuous iff its composite with $e_A$ is continuous. Now $u_A\circ e_A=p$ and $\rho_\alpha\circ e_A=r_\alpha$ by (3.4), and $p$ and $r_\alpha$ are continuous by hypothesis. Finally $e_A$ is a quotient map by construction. ◻

Proof. Theorem 5.1 supplies the unique function $\phi_{\mathcal E}\colon E\to E_A$ satisfying $e_A=\phi_{\mathcal E}\circ e$, and asserts its surjectivity. For continuity: the composite $\phi_{\mathcal E}\circ e=e_A$ is continuous (it is the quotient map onto $E_A$), and $e$ is a quotient map by Definition 10.1; by the characteristic property of quotient maps, a function out of $E$ whose composite with the quotient map $e$ is continuous is itself continuous. Hence $\phi_{\mathcal E}$ is continuous. Uniqueness among continuous morphisms is inherited from uniqueness among all set maps, proved in Theorem 5.1 (v). Finally $\mathcal E_A$ is itself an object of the category by Proposition 10.2. ◻

Proof. We have $\iota\circ e_A=j_A$, which is continuous; since $e_A$ is a quotient map, $\iota$ is continuous by the characteristic property. Continuity of $\iota$ says exactly that every subspace-open subset of $E_A$ (i.e. every $\iota^{-1}(W)$ with $W$ open in $T\times V_A$) is quotient-open; hence the quotient topology contains the subspace topology.

$\iota$ is injective and continuous, so it is an embedding iff it is open onto its image, i.e. iff every quotient-open subset of $E_A$ is subspace-open. Combined with , this holds iff the two topologies are equal.

For the reformulation, write $I:=\mathop{\mathrm{Im}}(j_A)$ with the subspace topology and let $\bar j_A\colon S\to I$ be the corestriction of $j_A$, so that as functions $\bar j_A=\iota\circ e_A$ with $\iota$ now regarded as the identity of underlying sets $E_A\to I$. Suppose the topologies agree. Then $\iota$ is a homeomorphism and $e_A$ is a quotient map, so $\bar j_A$ is a quotient map. Conversely, suppose $\bar j_A$ is a quotient map. A subset $U\subseteq E_A$ is quotient-open iff $e_A^{-1}(U)$ is open in $S$; and $e_A^{-1}(U)=\bar j_A^{-1}(U)$ as subsets of $S$, since $e_A$ and $\bar j_A$ agree as functions into the common underlying set. Because $\bar j_A$ is a quotient map, $\bar j_A^{-1}(U)$ open in $S$ implies $U$ open in $I$, i.e. subspace-open. Hence quotient-open implies subspace-open, and with the two topologies coincide. ◻

Proof. Write $I=\mathop{\mathrm{Im}}(j_A)\subseteq T\times V_A$ with the subspace topology, and let $\bar j_A\colon S\to I$ be the (continuous, surjective) corestriction of $j_A$. As a subspace of a Hausdorff space, $I$ is Hausdorff. We show $\bar j_A$ is a closed map, hence a quotient map, and then apply Proposition 10.4 .

Let $C\subseteq S$ be closed. Since $S$ is compact, $C$ is compact; since $\bar j_A$ is continuous, $\bar j_A(C)$ is a compact subset of $I$; since $I$ is Hausdorff, the compact subset $\bar j_A(C)$ is closed in $I$. Thus $\bar j_A$ is a continuous closed surjection, and every continuous closed surjection is a quotient map: if $\bar j_A^{-1}(U)$ is open then $S\setminus\bar j_A^{-1}(U)$ is closed, its image $\bar j_A(S\setminus\bar j_A^{-1}(U))=I\setminus U$ is closed by surjectivity, so $U$ is open. By Proposition 10.4, the topologies agree and $\iota$ is an embedding. Finally $E_A\cong I$ is compact (continuous image of a compact space) and Hausdorff (subspace of a Hausdorff space). ◻