Showing $U\otimes (V\oplus W) \cong U\otimes V \oplus U\otimes W $ I have to show $U\otimes (V\oplus W) \cong U\otimes V \oplus U\otimes W $ where $U,V,W$ are $K-$vector spaces. One way to give a linear map from left to right is: 
$$u\otimes (v,w)\mapsto (u\otimes v, u\otimes w).$$ This is well-defined since vectors of the form on the left span $U\otimes (V\oplus W)$. 
I cannot find the inverse of this map. Please give me some hints for this!
 A: You can define tensor product by the universal property. Then you just need to check that $U\otimes V\oplus U\otimes W$ satisfies the universal property for tensor product of $U$ and $V\oplus W$. This check is pretty much trivial. This approach is very useful, because it allows you to prove a lot of other similar identities, and not necessarily for finite dimensional spaces.
A: Your first map can be written as 
$$F\left(\sum_{i=1}^n u_i\otimes(v_i,w_i)\right)=\left(\sum_{i=1}^n u_i\otimes v_i,\sum_{i=1}^n u_i\otimes w_i\right)$$
Now the inverse is 
$$F^{-1}\left(\sum_{i=1}^n u_i\otimes v_i,\sum_{j=1}^m u'_j\otimes w_j\right)=\sum_{i=1}^n u_i\otimes (v_i,0)+\sum_{j=1}^m u'_j\otimes (0,w_j).$$
A: Hint: pick bases for $U,V,W$ to induce bases for $U\otimes(V\oplus W)$ and $(U\otimes V)\oplus (U\otimes W)$. How does this prove the obvious map is an isomorphism between them?
A: Here are some of the details implicit in @Sasha's answer.

Notation: Given vectors spaces $X,Y$ and elements $x \in X$, $y \in Y$, I'll write $x \oplus y$ for the corresponding element of the vectors space direct sum $X \oplus Y$. 

The map
\begin{align*}
 U \times  (V \oplus W) \to (U \otimes V) \oplus (U \otimes W) 
&&
(u, v \oplus w) \mapsto (u \otimes v ) \oplus (u \otimes w).
\end{align*}
is bilinear so, by the universality of the canonical bilinear map 
$$(u, v \oplus w) \mapsto u \otimes (v \oplus w) : U \times (V \oplus W) \to U \otimes (V \oplus W),$$ there is a unique linear map 
$$\Phi : U \otimes (V \oplus W) \to (U \otimes V) \oplus (U \otimes W)$$ 
such that 
\begin{align*}
\Phi(u \otimes (v \oplus w)) = (u \otimes v) \oplus (u \otimes w)
&& \text{ for all } u \in U, v \in V, w \in W.
\end{align*}
Similarly, 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, $\Phi_1$ and $\Phi_2$ 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*}
One then checks that $\Phi$ and $\Psi$ are inverse to one another (this can also be achieved using universal properties). 
