Is the following 3-manifold irreducible?

Question

We start with product manifold $ X := S_g \times (0,1)$, where $S_g$ is a closed, orientable surface of genus $g \geq 1$. Since this has $\mathbb R^3$ as its universal cover, $X$ is irreducible. Now $X$ can also be seen as the interior of $\bar{X} = S_g \times [0,1]$. We want to manipulate $X$ a bit in order to create a bordism between $S_g$ and $S_{g+1}$, whose interior is ,hopefully, also an irreducible manifold.

Consider a boundary component $R := S_g \times \{0\}$ and fix a 2-disc $D \subset R$. This disc has an open subset $U \subset \bar{X}$ that is diffeomorphic to $D \times [0,\epsilon)$, such that $U \cap R = D$ (This follows since $R$ has a collar). Choose two distinct points $p,q \in D$ and connect them via a simple (injective) smooth arc $\gamma \subset U$ with the property that $\gamma \cap R = \partial \gamma = \{p,q\}$. Moreover, to ensure that $\gamma$ is as "simple" as possible, we additionally require that there exists a simple path $\gamma' \subset D$ between $p$ and $q$ such that the simple closed curve $\gamma \cup \gamma'$ bounds a disc $D' \subset U$, intersectig $D$ only in its boundary arc $\gamma'$. Drawing a picture isn't hard and really helps to make this paragraph more clear.

Now consider a tubular neighborhood $N \subset U$ of $\gamma$, such that $N \cong D_2 \times [0,1]$, where $D_2$ denotes the standart open unit disc of $\mathbb R^2$ and define $\bar{Y} := \bar{X} \setminus N$. Effectively, we have removed an open half-tube from $\bar{X}$ and simultaneously added a handle to the boundary component $R$, thereby increasing its genus by $1$. Hence, $\bar{Y}$ describes a bordism between $S_g$ and $S_{g+1}$ and, up to diffeomorphism, it doesnt depend on the exact choices of $D,p,q,U$ and $\gamma $. If $Y$ denotes its interior, is it clear that $Y$ is still irreducible ?

Remarks:

1. If $\tilde{Y}$ denotes the universal cover of $Y$, then $Y$ is irreducible if and only if $H_2(\tilde{Y}) = 0$, which follows from Hurewicz and the sphere theorem.

2. Knowing that $\mathbb R^3$ covers $X$, what can be said about $\tilde{Y}$? Since $Y$ can be regarded as an open subset of $X$ and $\mathbb R^3$ covers $X$ via some map $p$, we know that each component of $p^{-1}(Y)$ is a covering space of $Y$. What can be said about those components ?

Answer 1

Yes, $Y$ is still irreducible, and this holds by a simple connectivity argument.

Suppose that $\Sigma \subset Y \subset X$ is a smoothly embedded 2-sphere. Since $X$ is irreducible, $\Sigma$ bounds a smoothly embedded ball $B \subset X$. Note that $\bar X - \Sigma$ has two components: the inside of $\Sigma$ which equals the interior of $B$, and the outside of $\Sigma$ which equals the complement of $B$. In $\bar X$, clearly $\Sigma$ is disjoint from $N$. Since $N \cup (S_g \times \{0\})$ is a connected subset of $\bar X - \Sigma$, and since $S_g \times \{0\}$ is in the outside of $\Sigma$, it follows that $N \cup (S_g \times \{0\})$ is in the outside of $\Sigma$. Thus $B \subset Y$.

Answer 2

To prove irreducibility it suffices to prove $\pi_2=0$. Assume $\pi_2Y\not=0$, then the sphere theorem gives you an embedded sphere representing a nontrivial element in $\pi_2Y$. we can assume that it doesn't hit the boundary of $Y$

Because of $\pi_2X=0$, this sphere bounds some map $f\colon B^3\to X$. After a small homotopy $f$ will be transversal to $\partial Y$, so $f^{-1}(X\setminus Y)$ is a properly embedded submanifold with boundary $f^{-1}(\partial Y)$.

But $f^{-1}(\partial Y)\cap S^2=\emptyset$, so the image of $f$ has to be contained in $Y$ and so the sphere does not represent a nontrivial element in $\pi_2Y$.