First some definitions: $$ f: V_1 \rightarrow V_2 , linear$$ $$ g: W_1 \rightarrow W_2 , linear$$
$$ h': V_1 \times W_1 \rightarrow V_2 \times W_2 $$
$$ h: V_1 \otimes W_1 \rightarrow V_2 \otimes W_2 , (v_1 \otimes w_1) \mapsto f(v_1) \otimes g(w_1) $$
$$ v_1 \in V_1, \quad v_2 \in V_2, \quad w_1 \in W_1, \quad w_2 \in W_2 $$
I was supposed to prove that there exists a unique linear map h (as defined) for all $$v \in V_1, \quad v \in W_1$$ I have already proven the existence of the linear map, however I am unsure if my proof that $$(v_1 \otimes w_1) \mapsto f(v_1) \otimes g(w_1) $$ is correct. Please take a look at it and let me know if something isn't right. Thank you!
proof: $$h(v_1 \otimes w_1) = v_2 \otimes w_2 \quad (1) $$ $$ v_2 \otimes w_2 = (\otimes \circ h')(v_1,w_1) = \otimes (h'(v_1,w_1)) = \otimes ((v_2,w_2)) = \otimes ((f(v_1),g(w_1))) = f(v_1) \otimes g(w_1) $$ therefore with (1) we have: $$ f(v_1) \otimes g(w_1) = h(v_1 \otimes w_1) $$
I used the tensor product symbol for both (the map and the operator). If it's infront of brackets, it's the map)