Euler characteristic of a cotangent bundle.

Question

Given a manifold $M$ with $\chi(M)=0$, I would like to prove that $\chi(T^{\ast}M)=0$. I know how to do it when $T^{\ast}M$ is globally trivial, i.e. $T^{\ast}M=\mathbb R^{n}\times M$, where $n$ is a dimension of $M$; but I am not sure how to do it in general. Any help will be highly appreciated.

Answer 1

If $X$ and $Y$ are homotopy equivalent spaces, then $H_i(X; \mathbb{Z}) \cong H_i(Y; \mathbb{Z})$ for all $i$. Therefore, if $X$ and $Y$ have well-defined Euler characteristics, they are equal.

Now note that if $E$ is the total space of a vector bundle over a space $B$, then $E$ and $B$ are homotopy equivalent ($E$ deformation retracts onto the zero section), so $\chi(E) = \chi(B)$ provided they are defined. In your case, $\chi(T^*M) = \chi(M) = 0$.