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](https://mathoverflow.net/questions/222651/is-the-following-3-manifold-irreducible) 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!



Constructing a "nice" cobordism
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. 


