# Constructing a “nice” cobordism

Denote by $\Sigma_g$ the closed, orientable surface of genus $g$. I want to construct a cobordism $M_g$ between $\Sigma_g$ and $\Sigma_{g+1}$ with the following two nice properties:

1) $M_g$ is an orientable Haken manifold

2) The two boundary inclusions $\Sigma_g \hookrightarrow M_g$ and $\Sigma_{g+1} \hookrightarrow M_g$ are both $\pi_1$-injective. (By the loop theorem, this is equivalent to both components of $\partial M_g$ being incompressible in $M_g$)

I have made some attemps at constructing such $M_g$, here is what I came up with. Consider the product $X_g := \Sigma_g \times [0,1]$ and a smooth curve $\gamma \subset X_g$, properly embedded and transverse to $\partial X_g$, so that $\partial \gamma \subset \Sigma_g \times \{0\}$. By removing from $X_g$ an open tubular neighborhood $N$ of $\gamma$ (and smoothing the corners), one obtains a cobordism $X_g^\gamma$ between $\Sigma_g$ and $\Sigma_{g+1}$. By slightly extending the argument given in the answer to this similar question I posed a while ago, one can show that $X_g^\gamma$ will indeed always by an orientable Haken manifold. However, there are examples of such $X_g^\gamma$, where the inclusion $\Sigma_{g+1} \hookrightarrow M_g$ is not $\pi_1$-injective, and I do not know if there even exists a loop $\gamma$ as above, such that the resulting space $X_g^\gamma$ will have two incompressible boundary components.

QUESTION: Can the construction explained above even be used to obtain a "nice" cobordism between $\Sigma_g$ and $\Sigma_{g+1}$ ? If not, can someone give me a hint on how to properly approach this particular problem ?

I do not expect fully fleshed out answers. I would be more than happy about some hints. Thanks in advance!

Yes, your construction works. Even with an "unknotted" arc your construction works. You can think of your manifold as a boundary connect sum of $S^1 \times S^1 \times [0,1] \setminus B^3$ with $\Sigma_g \times [0,1]$ along some embedded $D^3$ in the $S^2$-component of the first manifold's boundary. So this reduces checking $\pi_1$-injectivity to the $g=0$ case.