show that composition and tensor product commutes. 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?

Yes looks good!
A few nit-picks (actually just one now that I think of it):

*

*You say l.h.s. = ... resp. r.h.s. = ..., but this should be $(l.h.s.)(u\otimes v)=\dots$ resp. $(r.h.s.)(u\otimes v)=\dots$. The left-hand side (resp. right-hand side) is just the function. You show these functions are equal by showing equality on the (arbitrary) element $u\otimes v$.

