Boundary connected sum of manifolds I have two related questions about the boundary connected sum of manifolds with boundaries.
Let $T=S^1 \times S^1$ be a torus and let $X=T \times [0, 1]$ be the cylinder over the torus. Let $X'$ be a copy of the cylinder over the torus.
I call the boundary surface $T\times 0$ the inner torus and call $T\times 1$ the outer torus.
We construct a new manifold from $X$ and $X'$ using the boundary connected sum as follows.
First, we apply the boundary connected sum to the outer tori of $X$ and $X'$. Namely, we identify a disk in the outer torus of $X$ with a disk in the outer torus of $X'$ via an orientation reversing homeomorphism. Let us denote by $Y$ be the resulting manifold. The manifold $Y$ contains the copies of the inner boundaries of $X$ and $X'$.
Next, we apply the boundary connected sum to the those copies of inner boundaries in $Y$. We call the resulting manifold $Z$
The question is;
Is the manifold $Z$ homeomorphic to the cylinder over a 2-torus (genus 2 surface)?
By construction the boundaries are tori. But I am not sure what this manifold is.
Also, if this is not the cylinder over a torus, is there any similar connected sum operation that produce the cylinder over a torus?
 A: @DanielRust is correct; what you have constructed is just another compact manifold whose boundary is two genus-2 surfaces, not the cylinder on that surface.  Here's some more detail. 
$Z$ is homotopy equivalent to the two cylinders joined by line segments rather than disks.  More precisely, $(X \coprod X' \coprod I_0 \coprod I_1) / R$, where each $I_j$ is a unit interval and the relation $R$ glues $0 \in I_j$ to $(t, j) \in X$ and $1 \in I_j$ to $(t, j) \in X'$ (choosing some base point $t \in T$).  Using another homotopy equivalence to collapse $X$ and $X'$ to $T$ and $I_0$ to a point, we get $T \vee T \vee S^1$.
To show that $\Bbb{Z}^2 \ast \Bbb{Z}^2 \ast \Bbb{Z}$ is different from the fundamental group of the genus-2 surface, we'll need something like MO "Free splittings of one-relator groups".
Note that your construction works similarly if you replace $T$ by any manifold $M$, and some simpler examples may be helpful.  If you take $M$ to be $S^1$, the result is a twice-punctured torus; if you take $M$ to be $S^2$, the result appears to be a twice-punctured $S^2 \times S^1$.
