Showing $U\otimes (V\oplus W) \cong U\otimes V \oplus U\otimes W $

Question

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!

Answer 1

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 $$F^{-1}\left(\sum_{i=1}^n u_i\otimes v_i,\sum_{j=1}^m u'_j\otimes w_j\right)=\sum_{i=1}^n u_i\otimes (v_i,0)+\sum_{j=1}^m u'_j\otimes (0,w_j).$$

Answer 2

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?

Answer 3

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.

Answer 4

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).