= Solution
Write $V=T_p\mathcal M$ and $V^*$ for its <dual space>. A type-$(1,1)$ <tensor> is equivalently a <linear map> $A:V\to V$: the corresponding scalar-valued <multilinear map> is $(X,\eta)\mapsto\eta(A(X))$. Thus the composition defines
$$
C(X,\eta)=\eta\bigl(B(A(X))\bigr).
$$
It is linear in $X$ because $A$ and $B$ are <linear maps>, and linear in the <covector> $\eta$ by evaluation. This proves directly that \b[the composition is a tensor of type $(1,1)$].
Choose a basis $e_\alpha$ of the <tangent space> and its dual basis $e^\beta$. With the paper's component order $A(e_\alpha)=A_\alpha{}^\gamma e_\gamma$,
$$
B(A(e_\alpha))=A_\alpha{}^\gamma B_\gamma{}^\beta e_\beta,
\qquad
\boxed{C_\alpha{}^\beta=A_\alpha{}^\gamma B_\gamma{}^\beta.}
$$
Equivalently, in the usual output-first matrix notation, $C^\beta{}_\alpha=B^\beta{}_\gamma A^\gamma{}_\alpha$: \b[the matrix product is $BA$, not $AB$]. This <composition of mixed tensors> contracts one output index with one input index. Under the <tensor component transformation law>, the two Jacobian factors at that contracted index cancel, leaving exactly one factor for each free index.
Back to article page
