Here is the question I am trying to solve :
If $f$ and $f'$ [resp. $g$ and $g'$] are composable linear maps, show that $$(f' \circ f) \otimes (g' \circ g) = (f' \otimes g') \circ (f \otimes g)$$
Here is the information I know:
Let $f: U \to U'$ and $g: V \to V'$ be linear maps. We define their tensor product $f \otimes g: U \otimes V \to U' \otimes V'$ by
$$(f \otimes g)( u \otimes v) = f(u) \otimes g(v)$$ for all $u \in U$ and $v \in V.$
Here is my attempt for the solution:
\begin{align*}
l.h.s &= ((f' \circ f) \otimes (g' \circ g) ) (u \otimes v)\\
&= (f' \circ f) (u) \otimes (g' \circ g) (v)\\
&= f'(f(u)) \otimes g'(g(v)) \quad \quad \quad \quad \quad \text{(1)}
\end{align*}
Where the second equality is by definition of tensor product of linear maps and the third equality is by definition of composition.
\begin{align*}
r.h.s &= ((f' \otimes g') \circ (f \otimes g))(u \otimes v)\\
&= (f' \otimes g') ((f \otimes g)(u \otimes v))\\
&= (f' \otimes g') (f(u) \otimes g( v))\\
&= f'(f(u)) \otimes g'(g(v)) \quad \quad \quad \quad \quad \text{(2)}
\end{align*}
Where the second equality is by definition of composition and the third equality is by definition of tensor product of linear maps and the last equality is again by definition of tensor product of linear maps where $f': U' \to U''$ and $g': V' \to V''$.
Is my solution correct? if not, how can I correct it?
Best Answer
Yes looks good! A few nit-picks (actually just one now that I think of it):