Betti numbers of non-orientable $3$-manifolds

Question

Let $M^3$ be a compact $3$-manifold with boundary $\partial M$.

If $M$ is orientable, then it is known (see Lemma 3.5 here) that $2\dim(\ker(H_1(\partial M,\mathbb{Q})\rightarrow H_1(M,\mathbb{Q})))=\dim(H_1(\partial M,\mathbb{Q}))$. This results is also known as "half lives, half dies principle", and in this setting it follows easily by considering the long exact sequence of the pair $(M,\partial M)$ and by using Poincaré-Lefschetz duality.

I would like to know whether an analogous result holds in the non-orientable case.

More precisely, suppose $M$ and $\partial M$ are both non-orientable (and for simplicity, assume both $M$ and $\partial M$ connected). I would like to say that if $b_1(\partial M;\mathbb{Q})=2h+1$ then $\dim(\ker(H_1(\partial M,\mathbb{Q})\longrightarrow H_1(M,\mathbb{Q})))=h$ (or at least $\geq h$).

If we consider the long exact sequence of the pair $(M,\partial M)$, it is not difficult to see that the statement above is equivalent to $b_2(M,\partial M;\mathbb{Q})\geq b_1(M;\mathbb{Q})-1$, or $b_2(M;\mathbb{Q})\leq b_1(M,\partial M;\mathbb{Q})$.

While I was not able to prove it nor to construct a counterexample, I think it is a reasonable statement essentially for 3 reasons:

1. Even if it is not the case I am interested in, it is true if $M$ is non-orientable and $\partial M=\emptyset$, in the sense that $b_2(M;\mathbb{Q})=b_1(M;\mathbb{Q})-1$,
2. It is true if $M$ is a non-orientable handlebody,
3. If the condition is true for $(M,\partial M)$ and $(N,\partial N)$ then it is true also for $(M\natural N,\partial M\#\partial N)$.

Answer 1

In the nonorientable case the “half lives, half dies principle” has the following form: $$\dim \ker \left( H_{1}(\partial M;\mathbb{Q}) \to H_{1}(M;\mathbb{Q}) \right) + \dim \ker \left( H_{1}(\partial M;\mathbb{Q}^{w}) \to H_{1}(M;\mathbb{Q}^{w}) \right) = H_{1}(\partial M;\mathbb{Q}),$$ where $w \colon \pi_{1}(M) \to \mathbb{Z}/2$ denotes the orientation character.

You can see that by looking at the oriented double cover $N$ of $M$. If $T \colon N \to N$ denotes the deck transformation corresponding to the cover, then the fundamental class of the pair $(N,\partial N)$ satisfies $T_{\ast}[N,\partial N] = -[N, \partial N]$. Therefore, if we denote by $H_{\ast}(N;\mathbb{Q})^{\pm}$, the $(\pm1)$-eigenspace of $T_{\ast}$, then the cap product with the fundamental class exchanges eigenspaces. Therefore, the Poincaré-Lefschetz duality isomorphism exchanges the eigenspaces as well, i.e., $$(-) \cap [N,\partial N] \colon H^{k}(N;\mathbb{Q})^{\pm} \cong H_{3-k}(N,\partial N; \mathbb{Q})^{\mp}.$$ Furthermore, we can identify the $1$-eigenspace of the action of $T$ on (co)homology with the untwisted (co)homology of $M$ and the $(-1)$-eigenspace with the (co)homology of $M$ with coefficients in $\mathbb{Q}^{w}$.

Consequently, by twisted Poincaré-Lefschetz duality: $$\begin{align*}\ker \left( H_{1}(\partial M; \mathbb{Q}) \to H_1(M; \mathbb{Q}) \right) &\cong \operatorname{im}\left( H_2(M, \partial M; \mathbb{Q}) \to H_{1}(M; \mathbb{Q}) \right)\\ &\cong \operatorname{im} \left( H^{1}(M; \mathbb{Q}^w) \to H^{1}(\partial M; \mathbb{Q}^w) \right).\end{align*}$$ By the Universal Coefficient Theorem $$\operatorname{im}\left( H^{1}(M; \mathbb{Q}^w) \to H^{1}(\partial M; \mathbb{Q}^w) \right) \cong H_{1}(\partial M; \mathbb{Q}^{w}) / \ker\left( H_{1}(\partial M; \mathbb{Q}^{w}) \to H_1(M; \mathbb{Q}^{w}) \right),$$ as desired.