= Solution
Write $h_{ij}={}^{(3)}g_{ij}$ for the positive spatial <metric tensor> and $h^{ij}$ for its inverse. Expanding the <3+1 decomposition> gives $g_{00}=N^2-h_{ij}N^iN^j$, $g_{0i}=-h_{ij}N^j$ and $g_{ij}=-h_{ij}$. The inverse is
$$
\boxed{g^{00}=N^{-2},\qquad g^{0i}=-N^iN^{-2},\qquad
g^{ij}=-h^{ij}+N^iN^jN^{-2}.}
$$
For example, $g^{00}g_{00}+g^{0i}g_{i0}=1$, and $g^{00}g_{0j}+g^{0i}g_{ij}=0$. The spatial block similarly gives $g^{i0}g_{0j}+g^{ik}g_{kj}=\delta^i{}_j$. These checks determine all blocks without treating the spatial block alone as the inverse of the four-metric.
Raising the normal covector with this inverse <metric tensor> yields
$$
\boxed{n^\mu=(N^{-1},-N^iN^{-1}).}
$$
Take $N>0$ so it is future-pointing. Its norm is $n_\mu n^\mu=1$. The <spatial projection tensor> obeys $P^\mu{}_\nu n^\nu=0$, and multiplication gives
$$
P^\mu{}_\alpha P^\alpha{}_\nu
=\delta^\mu{}_\nu-2n^\mu n_\nu+n^\mu(n_\alpha n^\alpha)n_\nu
=\boxed{P^\mu{}_\nu}.
$$
Thus it is an <idempotent> <linear projection> onto vectors tangent to the spatial <hypersurface>. The negative spatial components of the spacetime <metric tensor> do not change this idempotence.
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