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Confirm $f \otimes g$ is a linear map

In the textbook, "Dancing with Qubits," it states:

"Let$V, W, X,$ and$Y$ be finite dimensional vector spaces over$\mathbb{F}$. If$f: V \rightarrow X$ and$g: W \rightarrow Y$ are linear maps, then so is$f\otimes g: V\otimes W \rightarrow X \otimes Y$ where we define$(f \otimes g)(v \otimes w) = f(v) \otimes g(w)$."

It asks me to confirm$f \otimes g$ is a linear map.

By the definition of a linear map, I need to show that for$a \in \mathbb{F}$ and$v_1 \otimes w_1$ and$v_2\otimes w_2$ in$V\otimes W$,

$(f \otimes g)(a(v\otimes w)) = a((f\otimes g)(v\otimes w)$

and

$(f\otimes g)(v_1 \otimes w_1 + v_2 \otimes w_2) = (f\otimes g)(v_1 \otimes w_1) + (f\otimes g)(v_2 \otimes w_2)$.

Proof. Let$V, W, X,$ and$Y$ be finite dimensional vector spaces over$\mathbb{F}$, let$f: V \rightarrow X$ and$g: W \rightarrow Y$ be linear maps, and let$a \in \mathbb{F}$. Then$(f \otimes g)(a(v\otimes w)) = (f\otimes g)((av) \otimes w) = f(av)\otimes g(w) = af(v) \otimes g(w) = a(f(v)\otimes g(w)) = a((f\otimes g)(v\otimes w)).$

Next let$v_1\otimes w_1$,$v_2 \otimes w_2 \in V\otimes W$. I get stuck here because I don't know how to say anything about$(f\otimes g)(v_1 \otimes w_1 + v_2 \otimes w_2)$ in terms of the expression$v_1 \otimes w_1 + v_2 \otimes w_2$.

Answer*

In order to prove $(f\otimes g)(v_1 \otimes w_1 + v_2 \otimes w_2) = (f\otimes g)(v_1 \otimes w_1) + (f\otimes g)(v_2 \otimes w_2)$ we must show why linearity holds. The best way to do this is to define a bilinear map $\varphi: V \times W \rightarrow V' \otimes W'$ such that $\varphi(v,w) := f(v) \otimes g(w)$. Checking that it is bilinear, we have, 
$$\varphi(v+v', w) = f(v+v') \otimes g(w) = (f(v) + f(v')) \otimes g(w) = f(v) \otimes g(w) + f(v') \otimes g(w) = \varphi(v, w) + \varphi(v',w)$$
and for a scalar $a \in \mathbb{F}$:
$$\varphi(av, w) = f(av) \otimes g(w) = af(v)\otimes g(w) = a\varphi(v,w)$$
and you can easily prove likewise linearity in the second argument. So $\varphi$ is bilinear. Then by the [universal property][1] of the tensor product, the bilinear map induces a unique linear map:
$$f\otimes g : V\otimes W \rightarrow V' \otimes W'$$ such that $$(f \otimes g) (v \otimes w) = \varphi(v,w) = f(v) \otimes g(w)$$. By this $f \otimes g$ is linear and you have:
$$(f \otimes g)(v_1 \otimes w_1 + v_2 \otimes w_2) = (f \otimes g)(v_1 \otimes w_1) + (f\otimes g)(v_2 \otimes w_2)$$ The linked Wikipedia page also shows how you may prove the above by using the fact that the tensor product of the vector spaces is a quotient space.


  [1]: https://en.wikipedia.org/wiki/Tensor_product#Universal_property

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