The following fairly specific question comes up in a bordism computation I'm trying to do:


Are there compact $\mathbb Z$-oriented $4k+2$ dimensional manifolds with boundary $M$ such that $im(H\_{2k+1}(M; \mathbb Z/2)\to H\_{2k+1}(M, \partial M; \mathbb Z/2))$ has odd dimension as a $\mathbb Z/2$ vector space?


Clearly the answer is no if $k=0$. Also, I can show that this can't happen if $\partial M=\emptyset$ using a combination of Poincare duality and the universal coefficient theorem. But I haven't been able to rule out the possibility if the boundary is non-empty, or to construct examples.


Thanks. 



Are there oriented $4k+2$ manifolds such that $im(H_{2k+1}(M; Z/2)\to H_{2k+1}(M, \partial M; Z/2))$ has odd dimension?
I claim it is not possible. The image is the rank of $H\_{2k+1}(M;\mathbb Z\_2)/rad$, where
$rad$ is the radical of the intersection form on $H\_{2k+1}(M;\mathbb Z\_2)$. 


The intersection form on $H\_{2k+1}(M;\mathbb Z\_2)/rad$ is hyperbolic, i.e. has a "symplectic" basis, therefore this vector space has even dimension.


Let me try to prove that it is hyperbolic: the tricky point is to show that all classes square to zero, i.e. $\langle x^2,[M,\partial M]\rangle =0$ for all $x\in H^{2k+1}(M,\partial M;\mathbb Z\_2) $.


Now $\langle x^2,[M,\partial M]\rangle =\langle \beta Sq^{2k}x,[M,\partial M]\rangle=\langle Sq^{2k}x,\beta [M,\partial M]\rangle=0$ where $\beta$ denotes the cohomology respectively homology Bockstein. 


