A torsion-free connection that is not metric-compatible

Define ${\mathrm{\nabla }}_{X}Y:={\sum }_{a,b,c}{X}^a{Y}^b{\mathrm{\Gamma }}^c{}_{ab}{\mathrm{\partial }}_c+{\sum }_{b}X(Y^b){\mathrm{\partial }}_b$ for smooth fields $X={\sum }_a{X}^a{\mathrm{\partial }}_a$, $Y={\sum }_b{Y}^b{\mathrm{\partial }}_b$ in the global chart, with the given coefficients. Then ${\mathrm{\nabla }}_{fX}Y=f{\mathrm{\nabla }}_{X}Y$ because the coefficients depend bilinearly on the components of $X,Y$, ${\mathrm{\nabla }}_X(fY)={\sum }_{a,b,c}{X}^a(fY{)}^b{\mathrm{\Gamma }}^c{}_{ab}{\mathrm{\partial }}_c+{\sum }_{b}X(f{Y}^b){\mathrm{\partial }}_b=f{\mathrm{\nabla }}_{X}Y+X(f)Y$ by the product rule, and $\mathbb{R}$-bilinearity holds componentwise; hence $\mathrm{\nabla }$ is an affine connection by [L1].

By [L2] the Christoffel symbols of $\mathrm{\nabla }$ in the global chart are the given ${\mathrm{\Gamma }}^0{}_{01}={\mathrm{\Gamma }}^0{}_{10}=1$ and ${\mathrm{\Gamma }}^a{}_{bc}=0$ otherwise; in particular ${\mathrm{\Gamma }}^a{}_{bc}={\mathrm{\Gamma }}^a{}_{cb}$ for all $a,b,c$, so $\mathrm{\nabla }$ is torsion-free by [L3].

The compatibility identity of [L3] fails at every point: for $(a,b,c)=(1,0,0)$ the left-hand side is ${\mathrm{\partial }}_1{A}_{00}=0$ because $A_{00}=-1$ is constant, while the right-hand side is ${\mathrm{\Gamma }}^0{}_{10}{A}_{00}+{\mathrm{\Gamma }}^0{}_{10}{A}_{00}=1\cdot (-1)+1\cdot (-1)=-2\mathrm{\ne }0$. By the equivalence [L3], $\mathrm{\nabla }$ is not compatible with $A$; it is therefore not metric-selected by [L4].

The connection of steps 1.1-1.2 is an affine connection that is torsion-free but not $A$-compatible, so it refutes the displayed statement. Since it is not compatible it also shows that the two conditions in the definition of a metric-selected connection are independent in the sense that torsion freedom alone does not force compatibility.