Relations between $K_3$ complex surface and $\mathbb{CP}^2$ What are the relations between $K_3$ complex surface and  $\mathbb{CP}^2$?
For one thing, I know that $K_3$ generates 4-dimensional bordism group $\Omega_4^{Spin}=\mathbb{Z},$
such that it has a signature $\sigma $ of $K_3$ is -16.
For another thing, I know that $\mathbb{CP}^2$ generates 4-dimensional bordism group $\Omega_4^{SO}=\mathbb{Z},$
such that it has a signature $\sigma $ of $\mathbb{CP}^2$ is 1.
Am I correct to claim that
$$
\bar{K_3} =\mathbb{CP}^2 \# ... \#  \mathbb{CP}^2 \text{ for 16 times of connected sums?}
$$
or
$$
{K_3} =\bar{\mathbb{CP}^2} \# ... \#  \bar{\mathbb{CP}^2} \text{ for 16 times of connected sums?}
$$
Here I consider $\sigma $ of $\bar K_3$ is +16. The $\sigma $ of $\bar {\mathbb{CP}^2}$ is -1.
If yes, how should we make the statements more rigorous? (or sketch a proof?)
 A: $K^3$ has a neat construction, it is somewhat easier to understand the analogous construction on the torus, though.
Consider $T^2\subset \mathbb{C}^2$. The action $(z,w)\mapsto (\overline{z},\overline{w})$ of conjugation has 4 fixed points. If we remove those fixed points, the action on the boundary circles is the antipodal map. $S^1$ quotiented out by the antipodal map is still $S^1$, so the quotient of the 4 punctured torus is the 4 punctured sphere. Then glue in 4 möbius bands.
We could have done this differently, though. We can remove the fixed points and glue the möbius bands in before we quotient out by conjugation. The antipodal map on the boundary circle extends over the möbius band and the quotient can be taken from there with the same resulting manifold.
The same thing is happening in 4 dimensions, except for $T^4$ you'll have 16 fixed points. So if you blow up 16 times before taking the quotient you'll get a manifold called $K^3$.
So, to stick with your notation,
$K = \left(T^4 \#\underbrace{\mathbb{CP}^2\#\dots\#\mathbb{CP}^2}_{\text{16 times}}\right)/(z \mapsto \overline{z})$
Viewing it this way, the remaining non-trivial homology classes come from tori that survive the quotient, although the basis can be formed entirely from spheres too.
