$\Lambda (A)$ as a pre-$\lambda$-ring

There exists an additive map $\lambda_S: \Lambda_{(T)}(A)\to \Lambda_{(S)}\Lambda_{(T)}(A)$ defined by

$$\lambda_S([a]_T)=[[a]_T]_{S} \qquad (\forall a \in A)$$

PROOF.. For $\alpha(T)=\prod_i (1-\xi_iT)$, we have

$$\sum_i [[ \xi_i]_T]_U$$

$$= \prod_i (1-[\xi_i]_T U))_W$$

$$= (\sum_n \sum_{i_1<i_2<\dots i_n} [\xi_{i_1}\dots \xi_{i_n}]_T(-U)^n)_W$$

$$= (\sum_n \sum_{i_1<i_2<\dots i_n} (1-\xi_{i_1}\dots \xi_{i_n}T)_W(-U)^n)_W$$

$$= (\sum_n (\prod_{i_1<i_2<\dots i_n} (1-\xi_{i_1}\dots \xi_{i_n}T))_W(-U)^n)_W$$

$$= (\sum_n (\sum_{j=0}^\infty L_{j,n}(a) T^j)_W(-U)^n)_W$$

So the required map is given by

$$(\sum_j a_j(T))_W \mapsto (\sum_n \sum_{j=0}^\infty ( L_{j,n}(a) T^j)_W (-U)^n)_W$$

$\qedsymbol$