In the following demonstration, there is an error, but I cannot find where. (I explicitely put the $c^2$ to keep track of units).
We consider a metric $g_{\mu\nu}$ with a signature $(-, +, +, +)$ :
$\boxed{g_{\mu\nu} = \begin{pmatrix} -c^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}} \qquad\qquad \boxed{ g^{\mu\nu} = \begin{pmatrix} \frac{-1}{c^2} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}} $
The tensor of a perfect fluid is: $ \boxed { \begin{aligned} T^{\mu\nu} &= \left(\rho c^2+P\right)u^{\mu}u^{\nu}+Pg^{\mu\nu}\\ T_{\nu}^{\mu} &= \left(\rho c^2+P\right)u^{\mu}u_{\nu}+P\delta_{\nu}^{\mu}\\ T_{\mu\nu} &= \left(\rho c^2+P\right)u_{\mu}u_{\nu}+Pg_{\mu\nu} \end{aligned} }$ where $\rho$ is the mass density.
We can easily check that $\rho c^2$ and $P$ have the same unit.
If we consider only the first component $00$, there is no pressure $P$ involved (I consider a cosmological case where the fluid is at rest in comoving coordinates). The only solution to have that is that $u^0u^0=-g^{00}$, so that the contravariant perfect fluid tensor become : $T^{00} = \left(\rho c^2+P\right)u^{0}u^{0}+Pg^{00}=\left(\rho c^2+P\right)\times{-g^{00}}+Pg^{00}=-\rho c^2 g^{00} = -\rho c^2 \times -\frac{1}{c^2}=\rho$ which is the expected result. So, to have the expected result, one should have $\boxed{\left(u^0\right)^2=+\frac{1}{c^2}}$.
Now come the weird part.
The metric can be written: $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}=-c^2dt^2+dx^2+dy^2+dz^2=dt^2\left(-c^2+\frac{dx^2}{dt^2}+\frac{dy^2}{dt^2}+\frac{dz^2}{dt^2}\right)\approx-c^2dt^2$ which is equivalent to :
$\frac{dt^2}{ds^2}\approx-\frac{1}{c^2}$
which means : $\boxed{\left(u^0\right)^2=-\frac{1}{c^2}}$
So we clearly have a problem of sign here between the two expressions of $u^0$.
QUESTION: where is the mistake ?