Find $(M,\partial M)$ with injective $H_k(\partial M)\rightarrow H_k(M)$. I am looking for  a compact $2k+1$-dimensional manifold $M$ ($k\ge1$) with boundary $\partial M$, such that
a) $H_k(\partial M;\mathbb{Q}) \neq 0$,
b) $\iota_*\colon H_k(\partial M;\mathbb{Q})\rightarrow H_k( M;\mathbb{Q})$ is injective. (Where $\iota\colon \partial M\hookrightarrow M$ is the inclusion).

I'm mainly interested in seeing whether a manifold like that even exists, so one particular example would be good enough. So far I have considered:


*

*$\partial M$ a closed surface of genus $g$, embedded into $\mathbb{R}^3$ in the usual way, such that it bounds a compact $3$-manifold $M$. For $g=0$, condition a) is not satisfied and for $g>0$ condition b) fails because $\dim H_1(M;\mathbb{Q})=g$ and $\dim H_1(\partial M;\mathbb{Q})=2g$.

*$M=X\times Y$, where $X$ is closed and even (resp. odd) dimensional and $Y$ has a boundary $\partial Y$ and is odd (resp. even) dimensional. Looking at the Künneth theorem it seems like injectivity of $H_i(\partial Y;\mathbb{Q})\rightarrow H_i( Y;\mathbb{Q})$ is required in all degrees $i$, so the problem becomes even harder (?). In order to find a low dimensional example I've considered $Y$ to be a bounded planar domain, but then again again b) fails for dimension reasons.

Questions: 


*

*How can I approach this problem and find an example for such an $M$? 

*Are there   any weird ways a surface can bound a $3$-manifold different from the above mentioned? 

*Is there  a table of manifolds with boundary and their homology that I can consider?

 A: For three dimensional manifolds there are no examples. Here I assume $M$ is orientable and $\partial M$ is connected. Then looking at the long exact sequence of the pair, we get
$$0\to H_2(M)\to H_2(M,\partial M)\to H_1(\partial M)\to H_1(M)\to H_1(M,\partial M)\to 0$$
Let $a$ be the rank of $H_2(M)$ and $b$ the rank of $H_2(M,\partial M)$. Then by Poincare-Lefschetz duality, $b$ is the rank of $H_1(M)$ and $a$ is the rank of $H_1(M,\partial M)$. Let $c$ be the rank of $H_1(\partial M)$. Then since the sequence is exact, the Euler characteristic is $0$, meaning $c=2b-2a$. Now if $H_1(\partial M)\to H_1(M)$ were injective, then $H_1(M)\cong H_1(M,\partial M)$ by the above sequence, implying $b=a$. But then $c=0$. I think you can generalize this to higher dimensions but I'm running short on time.
A: Cheerful Parsnips argument generalizes to higher dimensions: Again assume that $M$ is orientable and $\partial M$ is connected. Denote $a_i=\mathrm{rank}(H_i(\partial M))$, $b_i=\mathrm{rank}(H_i( M))$, $c_i=\mathrm{rank}(H_i(M,\partial M))$, then by Poincare duality for $\partial M$ we have $a_i=a_{n-1-i}$ and by Poincare-Lefschetz duality for $(M,\partial M)$ we have $b_{n-i}=c_{i}$.
Assuming that $H_k(\partial M)\rightarrow H_k(M)$ is injective, the long exact sequence for the pair $(M,\partial M)$ breaks up into two pieces and yields
$$
0=\sum_{i=0}^k (-1)^ i a_i - \sum_{i=0}^k (-1)^ i b_i + \sum_{i=0}^k (-1)^ i c_i
$$ 
and 
$$
0=\sum_{i=k+1}^{n-1} (-1)^ i a_i - \sum_{i=k+1}^n (-1)^ i b_i + \sum_{i=k+1}^n (-1)^ i c_i.
$$
Change indices in the latter equation and use the duality results to obtain
$$
0=\sum_{i=0}^{k-1} (-1)^ i a_i + \sum_{i=0}^k (-1)^ i c_i - \sum_{i=0}^k (-1)^ i b_i,
$$ 
where the sign changes are due to the fact that $n$ is odd and $n-1$ is even. Subtract this from the first equation, then 
$$
 a_k=0
$$
follows.

EDIT: In the comments it was claimed that $d:=\mathrm{rank}\left(\ker(H_k(\partial M)\rightarrow H_k(M)\right)=\frac{1}{2}\mathrm{rank}H_k(\partial M)$, this also follows by taking the long exact sequence for $(M,\partial M)$ and spitting it into:
$$\dots \rightarrow H_{k+1}(M,\partial M) \rightarrow\ker(H_k(\partial M)\rightarrow H_k(M)) \rightarrow 0 $$
and
$$
0\rightarrow \ker(H_k(\partial M)\rightarrow H_k(M))  \hookrightarrow H_k(\partial M)\rightarrow H_k(M) \rightarrow \dots
$$
and then doing essentially the same computations as above to obtain $a_k-2d=0$.
