As the headline says, I want to show that for any $R$-modules $U,\ V,\ W$ we have that $U\otimes(V\oplus W)\cong(U\otimes V)\oplus (U\otimes W)$. I know that there has been questions for this before, but my question has to do with one of the answers in this question.
Let $R$ be a commutative ring and consider the two $R$-modules $U$ and $V\oplus W$. We know that there exists a tensor product
$$
\lambda :U\times (V\oplus W) \rightarrow U\otimes (V\oplus W),
$$
which satisfies the universal property: for any bilinear transformation $\phi:U\times (V\oplus W)\rightarrow X$, where $X$ is any $R$-module, there exists a unique homomorphism $\xi :U\otimes (V\oplus W)\rightarrow X$ such that $\xi\ \circ\lambda=\phi$.
Consider $\phi :U\times (V\oplus W)\rightarrow (U\otimes V) \oplus (U\otimes W)$ given by $\phi(u,v\oplus w):=(u\otimes v)\oplus(u\otimes w)$. Then
$$
\xi :U\otimes (V\oplus W)\rightarrow (U\otimes V) \oplus (U\otimes W)
$$
is a homomorphism given by $\xi(u\otimes(v\oplus w)):=(u\otimes v)\oplus(u\otimes w)$. If I can find an inverse
$$
\xi^{-1}:(U\otimes V) \oplus (U\otimes W)\rightarrow U\otimes (V\oplus W)
$$
of $\xi$, then I'm done, but I am not sure how to find that. In the question I referred to earlier one of the answers wrote:
the maps \begin{align*} U \times V \to U \otimes (V \oplus W) && (u,v) \mapsto u \otimes (v \oplus 0) \\ U \times W \to U \otimes (V \oplus W) && (u,w) \mapsto u \otimes (0 \oplus w) \end{align*} are bilinear, and so give rise to linear maps \begin{align*} \Psi_1 : U \otimes V \to U \otimes (V \oplus W) \\ \Psi_2 : U \otimes W \to U \otimes (V \oplus W) \end{align*} such that \begin{align*} \Psi_1(u \otimes v) = u \otimes (v \oplus 0) && \text{ for all } u \in U, v \in V \\ \Psi_2(u \otimes w) = u \otimes (0 \oplus w) && \text{ for all } u \in U, w \in W. \end{align*} By the universal property of vector space direct sum, $\Psi_1$ and $\Psi_2$ can be combined into a linear map $$\Psi : (U \otimes V) \oplus (U \otimes W) \to U \otimes (V \oplus W)$$ such that \begin{align*} \Psi( (u \otimes v) \oplus (u' \otimes w)) = u \otimes (v \oplus 0) + u' \otimes (0 \oplus w) && \text{ for all }u,u' \in U, v \in V, w \in W. \end{align*}
and he says that we can identify $\Psi$ with $\xi^{-1}$. I understand how he obtains $\Psi_1$ and $\Psi_2$ from the universal property, but I don't see why we can combine them into $\Psi$. Can someone make that part more clear?