The question looks like an exercise in elementary algebraic topology, but I didn't manage to solve it. I am considering this question because it is a toy example in a problem I'm thinking about.


Let's consider the natural embedding $GL\_n(\mathbb C) \to \mathbb C^{n^2} \backslash \{0\}$. As was discussed [in this question](https://mathoverflow.net/questions/18677/cohomology-rings-of-gl-nc-sl-nc), cohomology with rational coefficients of $GL\_n(\mathbb C)$ is an exterior algebra on generators in degrees 1, 3, ..., 2n-1 (one generator in each degree), whereas $\mathbb C^{n^2} \backslash \{0\}$ is homotopy equivalent to a sphere $S^{n^2-1}$.


I'd like to prove that the map induced on cohomology of degree $n^2 - 1$ is a zero map. 
Any ideas?


Thanks.



Natural embedding GL_n(C) -> C^{n^2} \ {0} induces zero on cohomology
$\mathbb C^{n^2} \smallsetminus {0}$ is homotopy equivalent to $S^{2n^2-1}$, not $S^{n^2-1}$, and $\mathrm H^{2n^2-1}(GL\_n(\mathbb C)) = 0$.


