I would like to show that $\chi(M)=\#(M_{\triangle}, M_{\triangle})$ when $M$ is a compact oriented manifold, where $M_{\triangle}$ is the diagonal submanifold of $M\times M$. My definition of $\chi(M):=\#(M,Z(M),TM)$, where $Z(M)$ is the zero section of $TM$.
I belive the idea is to consider a tubular neighborhood $W$ of $M_{\triangle}$ and we have a canonical isomorphism between the normal bundle of $M_{\triangle}$ in $M$ and $TM$. So that we get an embedding $\phi :TM\rightarrow W$ such that $\phi|M=id$. Now take a section $f:M\rightarrow TM$ that is transverse to the zero section, and if we consider $\phi\circ f\circ i $, where $i: M_{\triangle}\rightarrow M$, we get that the map $\phi\circ f\circ i : M_{\triangle}\rightarrow M\times M$ is transverse to $M_{\triangle}$ and homotopic to the identity and so we can use it to compute the intersection number. Now it's clear that the set of points that intersect $M_{\triangle}$ are the points where $f$ vanishes but the part is troubling me is what does $\phi $ do to the orientation ? That is generally speaking if I have a tubular neighborhood and the normal bundle of the submanifold is orientable and so is the manifold of which we are making the tubular neighborhood will the tubular neighborhood map be orientation-preserving ? I know that to define this map we start by considering $x+y$ and of course this will be orientation-preserving but then we take on extensions and such and I don't know if this will ruin the orientability.
If it does happen to be that $\phi$ preserves the orientation then we have that $\chi(M)=\#(M_{\triangle},M_{\triangle})$, if it could reverse the orientation then I am out of ideas on how to prove this fact.
Any enlightment is appreciated. Thanks in advance.