Thom isomorphism theorem

For an $R$-oriented rank-$d$ vector bundle, multiplication by the Thom class gives isomorphisms

For a closed smooth submanifold $M\subseteq N$ of codimension $c>0$, a tubular neighborhood and the homological Thom isomorphism theorem identifySubstitution in the long exact sequence in relative homology gives the displayed embedding Gysin sequence. It does not require orientability over ${\mathbb{F}}_2$.

For a closed smooth embedding $i:N\hookrightarrow M$ of codimension $r$ whose normal bundle is oriented over a coefficient ring $R$, a tubular neighborhood, excision and the Thom isomorphism theorem identifyThe relative-to-absolute map defines $i_{!}:H^j(N;R)\to H^{j+r}(M;R)$. For closed oriented $M,N$, it is characterized bywith compatible orientations. It obeys $i_{!}(b⌣i^{\ast }c)=i_{!}b⌣c$. If $j:M\setminus N\hookrightarrow M$ and $\delta :H^q(M\setminus N;R)\to H^{q+1-r}(N;R)$ is the connecting map followed by the inverse Thom isomorphism, then, over ${\mathbb{F}}_2$,These identities follow from naturality of the relative cup product and show how the exact sequence determines multiplication in complements. Over ${\mathbb{F}}_2$ every real normal bundle has the required orientation.