A student asked me this. Suppose that $M$ is a compact, orientable $n$-manifold with boundary. It is a fact that for each $k$ with $0\leq k\leq n$ the vector spaces $H_k(M;\mathbb{R})$ and $H_{n-k}(M,\partial M;\mathbb{R})$ are isomorphic. Moreover, an isomorphism may be chosen between then such that in the diagram $$ \require{AMScd} \begin{CD} H_k(M;\mathbb{R}) @>{i}>> H_k(M,\partial M;\mathbb{R});\\ @V{\simeq}VV @V{\simeq}VV\\ H_{n-k}(M,\partial M;\mathbb{R}) @<{j}<< H_{n-k}(M;\mathbb{R}); \end{CD} $$ the inclusion induced maps $i$ and $j$ have matrix representatives that are transposes of each other. Now suppose that $\dim(M)=2m+1$. Prove that $\dim H_m(\partial M;\mathbb{R})$ is even, and that the map $H_m(\partial M;\mathbb{R})\rightarrow H_m(M;\mathbb{R})$ has rank equal to $\frac{1}{2}\dim H_m(\partial M;\mathbb{R})$.
I'm not sure what to do here. It seems if you let $n=2m+1$ and $k=m$ then $i$ must be injective due to the diagram. But, then $H_m(\partial M;\mathbb{R})\rightarrow H_m(M;\mathbb{R})$ would be the zero map by the sequence of the pair.
Consider this part of the long exact sequence for $\partial M\subset M$: $$H_{m+1}(M,\partial M;\mathbb R)\xrightarrow{a} H_m(\partial M;\mathbb R)\xrightarrow{b} H_m(M;\mathbb R).$$ This sequence is isomorphic to its dual $$H_m(M;\mathbb R)^*\xrightarrow{b^t} H_m(\partial M;\mathbb R)^*\xrightarrow{a^t} H_{m+1}(M,\partial M;\mathbb R)^*$$ (dual spaces, transposed maps) via Poincaré duality in each term (the isomorphism works for the entire long exact sequence, but we only need this part). Hence the kernel of $b$ is equal to the image of $a$, which is isomorphic to the image of $b^t$, which is equal to $(H_m(\partial M;\mathbb R)/(\ker b))^*$. As $\ker b\cong (H_m(\partial M;\mathbb R)/(\ker b))^*$, we get $\dim H_m(\partial M;\mathbb R)=2\dim\ker b$, as you wanted.
Notice though that I used more machinery than I see from your diagram, namely some properties of $a$ and $b$. So I don't know whether you'll find this answer satisfactory.