4. As shown in Figure 2, the diagonals $AC$ and $BD$ of square $ABCD$ intersect at point $M$, dividing the square into four triangles. $\odot O_{1}$, $\odot O_{2}$, $\odot O_{3}$, and $\odot O_{4}$ are the incircles of $\triangle AMB$, $\triangle BMC$, $\triangle CMD$, and $\triangle DMA$, respectively. Given that $AB=1$, the area of the central (shaded) part enclosed by $\odot O_{1}$, $\odot O_{2}$, $\odot O_{3}$, and $\odot O_{4}$ is ( ).
(A) $\frac{(4-\pi)(3-2 \sqrt{2})}{16}$
(B) $\frac{(3-2 \sqrt{2}) \pi}{4}$
(C) $\frac{(4-\pi)(3-2 \sqrt{2})}{4}$
(D) $\frac{4-\pi}{16}$
4.C.

Since $A C=B D=\sqrt{2}$, therefore, $A M=B M=\frac{\sqrt{2}}{2}$.
Let the inradius of the right triangle $\triangle A M B$ be $r$, then $\frac{\sqrt{2}}{2}-r+\frac{\sqrt{2}}{2}-r=1$, which means $r=\frac{\sqrt{2}-1}{2}$.
The area of the shaded part is equal to the area of the square $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ minus the area of the inscribed circle, thus
$$
\begin{array}{l}
S_{\text {ratio }}=4 r^{2}-\pi r^{2}=(4-\pi) r^{2} \\
=(4-\pi) \frac{(\sqrt{2}-1)^{2}}{4}=\frac{(4-\pi)(3-2 \sqrt{2})}{4} .
\end{array}
$$