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(1) I suppose that you are asking for an explicit construction of such a bordism. Its existence is already answered by your question; that $\mathbb{C}P^2$ generates $\Omega^{SO}_4$ just literally means any closed oriented smooth 4-manifold is oriented cobordant to some copies of $\mathbb{C}P$ or $\overline{\mathbb{C}P}^2$.

$H^2(K3,\mathbb{Z})$ is a rank-22 unimodular lattice and can be decomposed as $2\,\overline{E_8}\oplus 3\,H_2$. Here $\overline{E_8}$ means the negative $E_8$ lattice. $H_2$ means the rank-2 hyperbolic lattice. The $E_8$ lattice is not diagonalizable but stably diagonalizable, i.e., $E_8\oplus \overline{I}\simeq \overline{I}\oplus 8\,I$, where $I$ denotes the positive rank-1 unimodular lattice. Therefore,

$$H^2(K3,\mathbb{Z})\oplus2\,I\ \simeq\ \oplus 16\,\overline{I}\,\oplus 2\,I\,\oplus 3\,H_2\,.$$

According to Freedman's theorem, this means a homeomorphism between the two connected sums

$$K3\ \#\ 2\,\mathbb{C}P^2\ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ 3\,S^2\!\times\!S^2$$

because both sides are smooth. This probably fails to be a diffeomorphism. But a stabilization theorem of C.T.C. Wall says that we can always obtain a diffeomorphism by adding sufficiently many copies of $S^2\!\times\!S^2$. Namely, there is an integer $k$ such that we have a diffeomorphism

$$K3\ \#\ 2\,\mathbb{C}P^2\ \#\ k\,S^2\!\times\!S^2 \ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ (3\!+\!k)\,S^2\!\times\!S^2\,.$$

Now, considering the traces of the connected sums on both sides, we obtain an oriented bordism $W_1$ from $K3\coprod 2\,\mathbb{C}P^2\coprod k\,S^2\!\times\!S^2$ to $\coprod 16\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!k)\,S^2\!\times\!S^2$. Viewed in an alternative way, $W_1$ is an oriented bordism from $K3$ to

$$\coprod 18\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!2k)\,S^2\!\times\!S^2\,.$$

It is then a trivial task to find an oriented bordism $W_2$ from the above to $\coprod 16\,\overline{\mathbb{C}P^2}$. Hence the combination of $W_1$ and $W_2$ gives an oriented bordism you want.

You may find relevant materials in Alexandru Scorpan's book The Wild World of 4-manifolds and the references thereof. I don't know the minimal value of $k$ above, which might be just zero. Thank @MarcoGolla for pointing out another similar construction and sharing a useful reference.

(2) I don't quite understand the question. If you have a spin 5-manifold, it induces a spin structure on each connected component of its boundary.

[Note: This answer was updated according to @MarcoGolla's comment to avoid a potential circularity.]

(1) The existence of such a bordism is already answered by your question: "$\mathbb{CP}^2$ generates $\Omega^{SO}_4$" literally means any closed oriented smooth 4-manifold is oriented cobordant to some copies of $\mathbb{CP}^2$ or $\overline{\mathbb{CP}^2}$. So I suppose that you are asking for an explicit bordism.

As @MarcoGolla pointed out, there is a diffeomorphism

$$K3\mathbin\#\mathbb{CP}^2\ \simeq\ 4\mathbb{CP}^2\mathbin\#19\overline{\mathbb{CP}^2}\,,$$

by Exercise 8.3.4(d) in Gompf and Stipsicz's book 4-manifolds and Kirby Calculus. Tracing the connected sums on both sides, we obtain a compact oriented smooth 5-manifold $W_1$ with boundary

$$\partial W_1\ \simeq\ K3\amalg 20\mathbb{CP}^2\amalg 4\overline{\mathbb{CP}^2}\,.$$

Gluing $W_1$ with 4 copies of $\mathbb{CP}^2\times I$ along the boundary, we obtain a compact oriented smooth 5-manifold $W_2$ with boundary

$$\partial W_2\ \simeq\ K3\amalg 16\mathbb{CP}^2\,.$$

Hence $W_2$ is what you want.

In addition to 4-manifolds and Kirby Calculus mentioned above, Scorpan's book The Wild World of 4-manifolds is also a good place to learn about 4-manifolds.

(2) I don't quite understand the question. If you have a connected spin manifold, it induces a spin structure on each connected component of its boundary.

(1) I suppose that you are asking for an explicit construction of such a bordism. Its existence is already answered by your question; that $\mathbb{C}P^2$ generates $\Omega^{SO}_4$ just literally means any closed oriented smooth 4-manifold is oriented cobordant to some copies of $\mathbb{C}P$ or $\overline{\mathbb{C}P}^2$.

$H^2(K_3,\mathbb{Z})$ is a rank-22 unimodular lattice and can be decomposed as $2\,\overline{E_8}\oplus 3\,H_2$. Here $\overline{E_8}$ means the negative $E_8$ lattice. $H_2$ means the rank-2 hyperbolic lattice. The $E_8$ lattice is not diagonalizable but stably diagonalizable, i.e., $E_8\oplus \overline{I}\simeq \overline{I}\oplus 8\,I$, where $I$ denotes the positive rank-1 unimodular lattice. Therefore,

$$H^2(K_3,\mathbb{Z})\oplus2\,I\ \simeq\ \oplus 16\,\overline{I}\,\oplus 2\,I\,\oplus 3\,H_2\,.$$

According to Freedman's theorem, this means a homeomorphism between the two connected sums

$$K_3\ \#\ 2\,\mathbb{C}P^2\ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ 3\,S^2\!\times\!S^2$$

because both sides are smooth. This probably fails to be a diffeomorphism. But a stabilization theorem of C.T.C. Wall says that we can always obtain a diffeomorphism by adding sufficiently many copies of $S^2\!\times\!S^2$. Namely, there is an integer $k$ such that we have a diffeomorphism

$$K_3\ \#\ 2\,\mathbb{C}P^2\ \#\ k\,S^2\!\times\!S^2 \ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ (3\!+\!k)\,S^2\!\times\!S^2\,.$$

Now, considering the traces of the connected sums on both sides, we obtain an oriented bordism $W_1$ from $K_3\coprod 2\,\mathbb{C}P^2\coprod k\,S^2\!\times\!S^2$ to $\coprod 16\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!k)\,S^2\!\times\!S^2$. Viewed in an alternative way, $W_1$ is an oriented bordism from $K_3$ to

$$\coprod 18\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!2k)\,S^2\!\times\!S^2\,.$$

It is then a trivial task to find an oriented bordism $W_2$ from the above to $\coprod 16\,\overline{\mathbb{C}P^2}$. Hence the combination of $W_1$ and $W_2$ gives an oriented bordism you want.

You may find relevant materials in Alexandru Scorpan's book The Wild World of 4-manifolds and the references thereof. I don't know the minimal value of $k$ above, which might be just zero. Thank @MarcoGolla for pointing out another similar construction.

(2) I don't quite understand the question. If you have a spin 5-manifold, it induces a spin structure on each connected component of its boundary.

(1) I suppose that you are asking for an explicit construction of such a bordism. Its existence is already answered by your question; that $\mathbb{C}P^2$ generates $\Omega^{SO}_4$ just literally means any closed oriented smooth 4-manifold is oriented cobordant to some copies of $\mathbb{C}P$ or $\overline{\mathbb{C}P}^2$.

$H^2(K3,\mathbb{Z})$ is a rank-22 unimodular lattice and can be decomposed as $2\,\overline{E_8}\oplus 3\,H_2$. Here $\overline{E_8}$ means the negative $E_8$ lattice. $H_2$ means the rank-2 hyperbolic lattice. The $E_8$ lattice is not diagonalizable but stably diagonalizable, i.e., $E_8\oplus \overline{I}\simeq \overline{I}\oplus 8\,I$, where $I$ denotes the positive rank-1 unimodular lattice. Therefore,

$$H^2(K3,\mathbb{Z})\oplus2\,I\ \simeq\ \oplus 16\,\overline{I}\,\oplus 2\,I\,\oplus 3\,H_2\,.$$

According to Freedman's theorem, this means a homeomorphism between the two connected sums

$$K3\ \#\ 2\,\mathbb{C}P^2\ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ 3\,S^2\!\times\!S^2$$

because both sides are smooth. This probably fails to be a diffeomorphism. But a stabilization theorem of C.T.C. Wall says that we can always obtain a diffeomorphism by adding sufficiently many copies of $S^2\!\times\!S^2$. Namely, there is an integer $k$ such that we have a diffeomorphism

$$K3\ \#\ 2\,\mathbb{C}P^2\ \#\ k\,S^2\!\times\!S^2 \ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ (3\!+\!k)\,S^2\!\times\!S^2\,.$$

Now, considering the traces of the connected sums on both sides, we obtain an oriented bordism $W_1$ from $K3\coprod 2\,\mathbb{C}P^2\coprod k\,S^2\!\times\!S^2$ to $\coprod 16\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!k)\,S^2\!\times\!S^2$. Viewed in an alternative way, $W_1$ is an oriented bordism from $K3$ to

$$\coprod 18\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!2k)\,S^2\!\times\!S^2\,.$$

It is then a trivial task to find an oriented bordism $W_2$ from the above to $\coprod 16\,\overline{\mathbb{C}P^2}$. Hence the combination of $W_1$ and $W_2$ gives an oriented bordism you want.

You may find relevant materials in Alexandru Scorpan's book The Wild World of 4-manifolds and the references thereof. I don't know the minimal value of $k$ above, which might be just zero. Thank @MarcoGolla for pointing out another similar construction and sharing a useful reference.

(2) I don't quite understand the question. If you have a spin 5-manifold, it induces a spin structure on each connected component of its boundary.

(1) I suppose that you are asking for an explicit construction of such a bordism. Its existence is guaranteed by that $\mathbb{C}P^2$ generates $\Omega^{SO}_4$, which is essentially proved by Thom.

$H^2(K_3,\mathbb{Z})$ is a rank-22 unimodular lattice and can be decomposed as $2\,\overline{E_8}\oplus 3\,H_2$. Here $\overline{E_8}$ means the negative $E_8$ lattice. $H_2$ means the rank-2 hyperbolic lattice. The $E_8$ lattice is not diagonalizable but stably diagonalizable, i.e., $E_8\oplus \overline{I}\simeq \overline{I}\oplus 8\,I$, where $I$ denotes the positive rank-1 unimodular lattice. Therefore,

$$H^2(K_3,\mathbb{Z})\oplus2\,I\ \simeq\ \oplus 16\,\overline{I}\,\oplus 2\,I\,\oplus 3\,H_2\,.$$

According to Freedman's theorem, this means a homeomorphism between the two connected sums

$$K_3\ \#\ 2\,\mathbb{C}P^2\ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ 3\,S^2\!\times\!S^2$$

because both sides are smooth. This probably fails to be a diffeomorphism. But a stabilization theorem of C.T.C. Wall says that we can always obtain a diffeomorphism by adding sufficiently many copies of $S^2\!\times\!S^2$. Namely, there is an integer $k$ such that we have a diffeomorphism

$$K_3\ \#\ 2\,\mathbb{C}P^2\ \#\ k\,S^2\!\times\!S^2 \ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ (3\!+\!k)\,S^2\!\times\!S^2\,.$$

Now, considering the traces of the connected sums on both sides, we obtain an oriented bordism $W_1$ from $K_3\coprod 2\,\mathbb{C}P^2\coprod k\,S^2\!\times\!S^2$ to $\coprod 16\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!k)\,S^2\!\times\!S^2$. Viewed in an alternative way, $W_1$ is an oriented bordism from $K_3$ to

$$\coprod 18\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!2k)\,S^2\!\times\!S^2\,.$$

It is then a trivial task to find an oriented bordism $W_2$ from the above to $\coprod 16\,\overline{\mathbb{C}P^2}$. Hence the combination of $W_1$ and $W_2$ gives an oriented bordism you want.

You may find relevant materials in Alexandru Scorpan's book The Wild World of 4-manifolds and the references thereof. I don't know the minimal value of $k$ above, which might be just zero.

(2) I don't quite understand the question. If you have a spin 5-manifold, it induces a spin structure on each connected component of its boundary.

(1) I suppose that you are asking for an explicit construction of such a bordism. Its existence is already answered by your question; that $\mathbb{C}P^2$ generates $\Omega^{SO}_4$ just literally means any closed oriented smooth 4-manifold is oriented cobordant to some copies of $\mathbb{C}P$ or $\overline{\mathbb{C}P}^2$.

$H^2(K_3,\mathbb{Z})$ is a rank-22 unimodular lattice and can be decomposed as $2\,\overline{E_8}\oplus 3\,H_2$. Here $\overline{E_8}$ means the negative $E_8$ lattice. $H_2$ means the rank-2 hyperbolic lattice. The $E_8$ lattice is not diagonalizable but stably diagonalizable, i.e., $E_8\oplus \overline{I}\simeq \overline{I}\oplus 8\,I$, where $I$ denotes the positive rank-1 unimodular lattice. Therefore,

$$H^2(K_3,\mathbb{Z})\oplus2\,I\ \simeq\ \oplus 16\,\overline{I}\,\oplus 2\,I\,\oplus 3\,H_2\,.$$

According to Freedman's theorem, this means a homeomorphism between the two connected sums

$$K_3\ \#\ 2\,\mathbb{C}P^2\ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ 3\,S^2\!\times\!S^2$$

because both sides are smooth. This probably fails to be a diffeomorphism. But a stabilization theorem of C.T.C. Wall says that we can always obtain a diffeomorphism by adding sufficiently many copies of $S^2\!\times\!S^2$. Namely, there is an integer $k$ such that we have a diffeomorphism

$$K_3\ \#\ 2\,\mathbb{C}P^2\ \#\ k\,S^2\!\times\!S^2 \ \simeq\ \#\ 16\,\overline{\mathbb{C}P^2}\ \#\ 2\,\mathbb{C}P^2\ \#\ (3\!+\!k)\,S^2\!\times\!S^2\,.$$

Now, considering the traces of the connected sums on both sides, we obtain an oriented bordism $W_1$ from $K_3\coprod 2\,\mathbb{C}P^2\coprod k\,S^2\!\times\!S^2$ to $\coprod 16\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!k)\,S^2\!\times\!S^2$. Viewed in an alternative way, $W_1$ is an oriented bordism from $K_3$ to

$$\coprod 18\,\overline{\mathbb{C}P^2}\coprod 2\,\mathbb{C}P^2\coprod (3\!+\!2k)\,S^2\!\times\!S^2\,.$$

It is then a trivial task to find an oriented bordism $W_2$ from the above to $\coprod 16\,\overline{\mathbb{C}P^2}$. Hence the combination of $W_1$ and $W_2$ gives an oriented bordism you want.

You may find relevant materials in Alexandru Scorpan's book The Wild World of 4-manifolds and the references thereof. I don't know the minimal value of $k$ above, which might be just zero. Thank @MarcoGolla for pointing out another similar construction.

(2) I don't quite understand the question. If you have a spin 5-manifold, it induces a spin structure on each connected component of its boundary.