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? 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. 
 A: 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. 
A: Martin O's answer is very nice. So in an oriented $2n$-manifold with $n$ odd the mod $2$ self-intersection of any $n$-dimensional mod $2$ homology class is $0$. 
Looking for a more geometric explanation of that, or anyway an explanation with no Steenrod operations in sight, I came up with the following (which is also related to John Klein's comment): 
Let's assume that the given class is represented by an immersed $n$-manifold $M$. The mod $2$ self-intersection number is then the evaluation on the mod $2$ fundamental class of $M$ of the mod $2$ Euler class of the normal bundle of the immersion. So it comes down to the following:
Claim: Let $n$ be odd and suppose that $M$ is a closed $n$-dimensional manifold and $E$ is a rank $n$ vector bundle such that the total space of $E$, considered as a $2n$-manifold, is orientable. Then the mod $2$ Euler class of $E$ is $0\in H^n(M;\mathbb Z/2)$. 
Proof: A rank $n$ vector bundle has a twisted integral Euler class, which belongs to $H^n(M;\Gamma)$, where $\Gamma$ is the coefficient system (locally isomorphic to $\mathbb Z$) associated with $w_1(E)$, the obstruction to orientability of $E$. The mod $2$ Euler class is the mod $2$ reduction of this, so it suffices if this twisted integral class is $0$. The (twisted) integral Euler class of a vector bundle of odd rank is always killed by $2$ (this is a standard fact in the oriented case, and it seems clear in the twisted case, too), so it suffices if the group $H^n(M;\Gamma)$ is torsion-free. But by Poincare duality it is isomorphic to $H_0(M;\mathbb Z)$, since $E$ and the tangent bundle of $M$ have the same orientability obstruction by hypothesis.
