Computing the euler Characteristic of $ S^2$ After defining the euler charateristic of a manifold $M$ as the intersection number of $M$ with it's tangent bundle , $\#(M,M,TM)$, Hirsch gives an example of computing for the special case where $M=S^n$.  So let $P$ denote the norh pole and $Q=-P$ the south pole. Let $\sigma: S^n-P\rightarrow \mathbb{R}^n$ and $\tau : S^n-Q\rightarrow \mathbb{R}^n$ be stereographic projections. Let $f$ be the vector field whose representation via $\sigma$ is the identity , I think what he means by this is that $T\sigma \circ f\circ \sigma^{-1}=Id:\mathbb{R}^n \rightarrow \mathbb{R}^n$. He claims that $f(x)\rightarrow 0$ as $x\rightarrow P$, and this I am not seeing why it's true. So let's suppose that $x\rightarrow P$ then we get that $\sigma(x)\rightarrow \infty$ and so we get that $T\sigma\circ f\circ \sigma^{-1} \rightarrow \infty$ but I don't see how we get that $f(x)\rightarrow 0$. I am not interaly sure of the affect of $T\sigma $ but I would say that $x\rightarrow P$ then $T_x\sigma (v)\rightarrow \infty$.
Then he defines the extension of $f$ to $S^n$ to be $f(P)=0$ and $f$ otherwise. Then in $\tau$ coordinates $f$ corresponds to $x\rightarrow -x$. And then he claims that $Ind_pf =1$ and $Ind_Qf=(-1)^n$. Shouldn't it be the other way around ? It doens't change the end result but I wanna make sure this makes sense or not.
Thanks in advance.
 A: In the $S^2$ case, you can gain intuition about the norm of $f(x)$ by considering circles centered around the origin of the plane of projection, which correspond to latitudinal circles on the sphere. In the plane, the circumference of such a circle is proportional to the norm of the identity vector field along it (under the usual metric on $\mathbb{R}^2$), but the corresponding latitudinal circle gets smaller as you go near the north pole and eventually reaches zero. This shows that the pullback metric from the sphere should decrease to zero faster than the identity vector field grows. This argument doesn't generalize very neatly to $S^n$, but a bit of computation can prove the point more concretely.
You can calculate the limit of $|f(x)|$ as $x\to P$ either in stereographic coordinates, or by pushing the identity vector field forward. Going the former route, the metric is (taken from Lee's textbook on Riemannian manifolds)
$$
\frac{4}{(1+ \sum u_i^2 )^2} \sum (d u_i)^2\ ,
$$
so the norm of $f(x)$ is $2r/(1+r^2)$ (where $r^2 = \sum u_i^2$), which approaches $0$ as $r\to\infty$.
In the $S^2$ case, the pushforward is not very hard either, where $f$ turns out to be $(-x z, -y z, 1-z^2)$, which approaches $0$ as $(x,y,z)\to (0,0,1)$.
Also I agree that the indices of $f$ at the poles should be the other way around.
