6. Let the odd function $f(x)$ have a domain of $[-2,2]$,

and be decreasing in the interval $[-2,0]$, satisfying
$$
f(1-m)+f\left(1-m^{2}\right)<0 \text {. }
$$

Then the range of real numbers $m$ is
6. $[-1,1)$.

Given that the domain of $f(x)$ is $[-2,2]$, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
-2 \leqslant 1-m \leqslant 2, \\
-2 \leqslant 1-m^{2} \leqslant 2
\end{array}\right. \\
\Rightarrow-1 \leqslant m \leqslant \sqrt{3} .
\end{array}
$$

Since $f(x)$ is an odd function and is decreasing in the interval $[-2,0]$, it follows that $f(x)$ is decreasing in the interval $[-2,2]$.
$$
\begin{array}{l}
\text { Therefore, } f(1-m)m^{2}-1 \\
\Rightarrow-2<m<1 .
\end{array}
$$

From equations (1) and (2), we know that $-1 \leqslant m<1$, which means the range of real numbers $m$ is $[-1,1)$.