Thom diagonal and zero-section collapse

For a metric bundle $\xi \to B$, define on the disk bundle $v⟼\pi (v)\wedge [v]\in B_{+}\wedge \mathrm{Th}(\xi )\mathrm{.}$ Every $v\in S(\xi )$ maps to the smash-product basepoint, independently of its base coordinate. The quotient universal property therefore gives the Thom diagonal ${\delta }_{\xi }:\mathrm{Th}(\xi )⟶B_{+}\wedge \mathrm{Th}(\xi );[v]⟼\pi (v)\wedge [v]\mathrm{.}$

The zero section is $s:B\to D(\xi )$, $s(b)=0_b$. It is continuous in every vector-bundle chart and is a section of $\pi$.

If an embedding $i:B\hookrightarrow M$ is supplied with tubular data consisting of an embedding $e:D(\xi )\to M$ satisfying $e\circ s=i$ and that is a homeomorphism onto a closed neighborhood $K$, carries the interior of $D(\xi )$ onto an open neighborhood $N$ of $B$, and carries $S(\xi )$ onto $K\setminus N$, the associated collapse is the based map $c:M_{+}⟶\mathrm{Th}(\xi )$ that sends $x\in K$ to $[e^{-1}(x)]$ and sends $M\setminus N$ and the disjoint basepoint to the Thom basepoint. On $K\setminus N$ the two formulas agree because $e^{-1}(x)\in S(\xi )$, so closed pasting makes the displayed map continuous. This is a definition conditional on supplied tubular data; no tubular-neighborhood existence theorem is asserted.

For an empty base the Thom diagonal is the unique based map. In rank zero it is the ordinary based diagonal $B_{+}\to B_{+}\wedge B_{+}$. Sphere points, the complement of the tubular neighborhood, and all quotient basepoints map to the stated basepoint. Identity bundle charts and the zero vector give the literal formulas. All maps are explicit and choice-free.