I am trying to review my differential Geometry. In J.Lee's Smooth Manifolds, there is an exercise in which one has to show that the product of smooth covering maps is a smooth covering map.
A smooth covering map is a smooth cover \pi: M' \rightarrow M, where M, \, M' are smooth manifolds, and in which every $p$ in M' has a neighborhood $U_p$ in $M$ such that $p|_{p^{-1}} (U_p) \rightarrow U_p$ (i.e., the restriction of $p$ to the inverse image of $U_p$ is a diffeomorphism) . I think this exercise is straightforward, (basically we just use the fact that the product of diffeomorphisms is a diffeomorphism) but I am kind of rusty, and would appreciate your inputs.
Now, we need to show that , given covers p_1:M' \rightarrow M and p_2:N' \rightarrow {N} that for any pair $(p,q)$ there is a neighborhood $W$ of $(p, q)$ such that $p_1 \times p_2 |_{p_1^{-1} \times p_2^{-1} (W)} : p_1^{-1} \times p_2^{-1} (W) \rightarrow W$ is a diffeomorphism.
But we know that for $p$ in $M$ there is a $U_p$ with $p_1|_{p_1^{-1} (U_p)} : p_1^{-1} (U_p) \rightarrow U_p$ and for any $q$ in $N$ there is a $U_q$ with $p_2|_{{p_2}^{-1} (U_q)} : p_2^{-1} (U_q) \rightarrow U_q$ such that both are diffeomorphisms. Then the product map is a diffeomorphism automatically, isn't it, i.e., isn't the map:
$(p_1, p_2)|_{(p_1^{-1} \times p_2^{-1}) (U_q)} : (p_1^{-1} \times p_2^{-1}) (U_q) \rightarrow U_p \times U_q$ a diffeomorphism?
We know that if $f = (f_1(x_1,\ldots,x_n),f_2(x_1,\ldots,x_n))$ is differentiable with derivative (f_1',f_2') and$f_1^{-1}$ and $f_2^{-1}$ are each differentiable (by the assumption of diffeomorphism), then I think the differentiable inverse here is given by the identities fof-1=Id.
I imagine there may be some issue in showing that the above is true for manifolds, and not just for subsets of $\mathbb{R}^n$. Other than that, is my setup correct?
Basically, I am curious as to whether this problem comes down to the fact that the product of diffeomorphisms is a diffeomorphism.
Thanks.