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I would like to give the following object, $M=S^5 \setminus \sqcup_{2 \text{ copies}} \text{int}(S^1\times D^4)$, a handle decomposition. It is then to be attached to another manifold. along the two copies of $S^1\times S^3$, so there is no need to start with a $0$-handle. My attempt, below, seems quite complicated (then I would like to generalize the construction to $S^5$ with more copies of $S^1\times D^4$ removed, under certain combinatorial conditions).

Is the following attempt a valid handle decomposition of $M$ ?

Attempt to construct M: Start with two free floating $1$-handles $(D^1\times D^4, S^0\times D^4)$. They have co-attaching region $\text{co-att}=D^1\times S^3$. I would think that I am not allowed to decompose $D^4=D^1\times D^3$ and connect two $D^4$ of each handle in order to create a $S^1\times S^3$. In that case only connect the two handles on one side to create a $S^1\times D^3$ and on the other ends as well. Take the result, which has boundary $(\sqcup_{2 copies} S^1\times D^3)\cup S^1 \times S^3$ and another copy of itself, and glue the four disjoint $S^1\times D^3$'s two by two along $S^2$'s, giving an object with two disjoint $S^1\times S^3$ as boundaries. There are also two unwanted boundaries from the co- attaching regions $\text{co-att}$, which are $\#_{2 copies} S^1\times S^3$. Fill part of this two spaces by attaching two $2$-handles $(D^2\times D^3,S^1\times D^3)$. The result are two copies of $D^4$, one for each "holes". Now the fundamental group is trivial, and we can close the remaining unwanted boundary with a 5-handle $(D^5\times D^0,S^4\times D^0)$.

I would like to give the following object, $M=S^5 \setminus \sqcup_{2 \text{ copies}} \text{int}(S^1\times D^4)$, a handle decomposition. It is then to be attached to another manifold. along the two copies of $S^1\times S^3$, so there is no need to start with a $0$-handle. My attempt, below, seems quite complicated (then I would like to generalize the construction to $S^5$ with more copies of $S^1\times D^4$ removed, under certain combinatorial conditions).

Is the following attempt a valid handle decomposition of $M$ ?

Attempt to construct M: Start with two free floating $1$-handles $(D^1\times D^4, S^0\times D^4)$. They have co-attaching region $\text{co-att}=D^1\times S^3$. I believe that I am not allowed to decompose $D^4=D^1\times D^3$ and connect two $D^4$ of each handle in order to create a $S^1\times S^3$. In that case, take one $D^4=D^1\times D^3$ of each handle and connect the $D_1$ of each two handles on one side to create a $S^1\times D^3$. Do the same on the other ends of the two 1-handles as well. The $D^1$'s of the co-attaching regions get connected as well to give a $S^1$. The result has boundary $(\sqcup_{2 copies} S^1\times D^3)\cup S^1 \times S^3$. Take the result and an another copy of itself, and connect the four disjoint $S^1\times D^3$'s two by two, giving an object with two disjoint $S^1\times S^3$ as boundaries; these are the boundaries of the target manifold. Note that at this point we did not do any gluings of handles but only combination of disks along their boundaries. There are also two unwanted boundaries from the co- attaching regions $\text{co-att}$, which are $\#_{2 copies} S^1\times S^3$. Fill part of this two spaces by attaching two $2$-handles $(D^2\times D^3,S^1\times D^3)$. The result are two copies of $D^4$, one for each "holes". Now the fundamental group is trivial, and we can close the remaining unwanted boundary with a 5-handle $(D^5\times D^0,S^4\times D^0)$.

I would like to give the following object, $M=S^5 \setminus \sqcup_{2 \text{ copies}} \text{int}(S^1\times D^4)$, a handle decomposition. It is then to be attached to another manifold. along the two copies of $S^1\times S^3$, so there is no need to start with a $0$-handle. My attempt seems quite complicated. Then I will generalize to $S^5$ with more copies of $S^1\times D^4$ removed, under certain combinatorial conditions. I lack an understanding of the rules for creating such object, so I am asking you if the following proposition is valid, or what you would do instead.

Start with two free floating $1$-handles $(D^1\times D^4, S^0\times D^4)$. They have co-attaching region $\text{co-att}=D^1\times S^3$. I would think that I am not allowed to decompose $D^4=D^1\times D^3$ and connect two $D^4$ of each handle in order to create a $S^1\times S^3$. In that case only connect the two handles on one side to create a $S^1\times D^3$ and on the other ends as well. Take the result, which has boundary $(\sqcup_{2 copies} S^1\times D^3)\cup S^1 \times S^3$ and another copy of itself, and glue the four disjoint $S^1\times D^3$'s two by two along $S^2$'s, giving an object with two disjoint $S^1\times S^3$ as boundaries. There are also two unwanted boundaries from the co- attaching regions $\text{co-att}$, which are $\#_{2 copies} S^1\times S^3$. Fill part of this two spaces by attaching two $2$-handles $(D^2\times D^3,S^1\times D^3)$. The result are two copies of $D^4$, one for each "holes". Now the fundamental group is trivial, and we can close the remaining unwanted boundary with a 5-handle $(D^5\times D^0,S^4\times D^0)$.

I would like to give the following object, $M=S^5 \setminus \sqcup_{2 \text{ copies}} \text{int}(S^1\times D^4)$, a handle decomposition. It is then to be attached to another manifold. along the two copies of $S^1\times S^3$, so there is no need to start with a $0$-handle. My attempt, below, seems quite complicated (then I would like to generalize the construction to $S^5$ with more copies of $S^1\times D^4$ removed, under certain combinatorial conditions).

Is the following attempt a valid handle decomposition of $M$ ?

Attempt to construct M: Start with two free floating $1$-handles $(D^1\times D^4, S^0\times D^4)$. They have co-attaching region $\text{co-att}=D^1\times S^3$. I would think that I am not allowed to decompose $D^4=D^1\times D^3$ and connect two $D^4$ of each handle in order to create a $S^1\times S^3$. In that case only connect the two handles on one side to create a $S^1\times D^3$ and on the other ends as well. Take the result, which has boundary $(\sqcup_{2 copies} S^1\times D^3)\cup S^1 \times S^3$ and another copy of itself, and glue the four disjoint $S^1\times D^3$'s two by two along $S^2$'s, giving an object with two disjoint $S^1\times S^3$ as boundaries. There are also two unwanted boundaries from the co- attaching regions $\text{co-att}$, which are $\#_{2 copies} S^1\times S^3$. Fill part of this two spaces by attaching two $2$-handles $(D^2\times D^3,S^1\times D^3)$. The result are two copies of $D^4$, one for each "holes". Now the fundamental group is trivial, and we can close the remaining unwanted boundary with a 5-handle $(D^5\times D^0,S^4\times D^0)$.