Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Milnor's statement is: "Let $M^r$ and $N^s$ be sub-manifolds of $V^{r+s}$ which are all smooth, compact, oriented and without boundary. If $p$ is a point of $M^r$ contained in an $r$-cell $U$, naturality of the Thom isomorphism implies that the inclusion induced map

$H_r(U,U-p) \longrightarrow H_r(V,V-N)$

is an isomorphism given by $\gamma \rightarrow \epsilon \psi(\alpha)$ where $\gamma$ is the orientation generator of $H_r(U,U-p)$, $\psi:H_0(N) \longrightarrow H_r(V,V-N)$ is the Thom isomorphism, $\alpha$ is the canonical generator of $H_0(N)$ and $\epsilon$ is the intersection number of $M$ and $N$ at $p$."

I understand that naturality of the Thom isomorphism is the fact that this diagram commutes, \begin{matrix} H_0(p)&\stackrel{j_*}{\longrightarrow}&H_0(N)\\ \downarrow{\psi}&&\downarrow{\psi}\\ H_r(U,U-p)&\stackrel{i_*}{\rightarrow}&H_r(V,V-N) \end{matrix}

where $i_*$ and $j_*$ are induced by inclusion. By the commutativity of the diagram, it is clear to me that $i_*$ is an isomorphism, but I don't see why $i_*(\gamma)=\epsilon \psi(\alpha)$.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.