Trivial Thom spaces as suspension smash products

With the product metric and the supplied trivialization, [F1] identifies the disk/sphere pair of $B\times {\mathbb{R}}^r$ with $(B\times D^r,B\times S^{r-1})$ and computes the quotient as $B_{+}\wedge (D^r\mathrm{/}{S}^{r-1})$, naturally in $B$. Under Disk bundle, sphere bundle, and Thom space: the differential topology interface this quotient is exactly $\mathrm{Th}(B\times {\mathbb{R}}^r)$, with the same based convention and the same compactly generated quotient topology.

Since $D^r\mathrm{/}{S}^{r-1}=S^r$ — with the conventions $S^{-1}=\mathrm{\varnothing }$ and $D^0\mathrm{/}{S}^{-1}=\{\mathrm{p}\mathrm{t}{\}}_{+}=S^0$ when $r=0$ — step 1.1 gives $\mathrm{Th}(B\times {\mathbb{R}}^r)\cong B_{+}\wedge S^r={\mathrm{\Sigma }}^r{B}_{+}$. For $r=0$ the empty sphere bundle and the convention $X\mathrm{/}\mathrm{\varnothing }=X_{+}$ leave $B_{+}$; for empty $B$ both sides are the one-point based space; for $r=1$ the boundary is the two endpoints. The identity formula commutes with pullback along every map $B^{\mathrm{\prime }}\to B$, which is the asserted naturality. This is the AT result restated for the framed DT target.