# Product of Smooth Covering Maps a Smooth Covering Map? ( J.Lee, 2-12)

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.

• You're never going to get anyone to read that if you don't TeX it. – Uticensis Mar 2 '11 at 6:58
• Sorry, it's my first post, I did not find the reference for TeX,but I will look it up now and will rewrite it; sorry. – user7679 Mar 2 '11 at 7:00
• Hope this is better, sorry. – user7679 Mar 2 '11 at 7:31
• Hey, don't apologize, I'm just trying to help you get your question answered. I formatted your question into a cleaner style, but I don't know your subject, so it's quite likely I butchered some of what you are trying to ask. However, I think I made it into a presentable start; I think if you edit my TeX job, you can get the hang of the notation and can fix whatever I screwed up. Good luck. – Uticensis Mar 2 '11 at 8:12
• I tried to fix some of the notation but it still isn't optimal you'll have to make some improvements yourself. – t.b. Mar 2 '11 at 8:29

## 1 Answer

You're right. To prove this you may use the fact that you have an atlas for $M_1\times M_2$ (it's the assumed differential structure of the exercise) whose elements are of the form $(U_1\times U_2,\varphi_1\times\varphi_2)$, with $(U_1,\varphi_1)$ being in an atlas for $M_1$ and $(U_2,\varphi_2)$ being in an atlas for $M_2$.

In fact, given the diffeomorphisms $f_1:M_1\to M_1'$ and $f_2:M_2\to M_2'$ you have for every $(p_1,p_2)\in M_1\times M_2$ smooth coordinate charts $(U_1\times U_2,\varphi_1\times\varphi_2)$ and $(V_1\times V_2,\psi_1\times\psi_2)$, with $(p_1,p_2)\in U_1\times U_2$ and $(f_1(p_1),f_2(p_2))=(q_1,q_2)\in V_1\times V_2$. So, the smoothnes of $$(\psi_1\times\psi_2)\circ(f_1\times f_2)\circ(\varphi_1\times\varphi_2)^{-1}= (\psi_1\circ f_1\circ\varphi_1^{-1})\times(\psi_2\circ f_2\circ\varphi_2^{-1})$$ and $$(\varphi_1\times\varphi_2)\circ(f_1\times f_2)^{-1}\circ(\psi_1\times\psi_2)^{-1} = (\varphi_1\circ f_1^{-1}\circ\psi_1^{-1})\times(\varphi_2\circ f_2^{-1}\circ\psi_2^{-1})$$ come from the smoothness of $$\psi_1\circ f_1\circ\varphi_1^{-1},$$ $$\psi_2\circ f_2\circ\varphi_2^{-1},$$ $$\varphi_1\circ f_1^{-1}\circ\psi_1^{-1},$$ and $$\varphi_2\circ f_2^{-1}\circ\psi_2^{-1}$$ as in Euclidean spaces.

• I think that this can be made into a complete argument, but you have to be a bit more careful: The maps $f_{1}$ and $f_{2}$ are not diffeomorphisms. You should start with neighborhoods $U_p$ and $U_{q}$ such that $f_{1}^{-1}(U_p)$ can be written as a disjoint union $U_{p}^i$ of open sets diffeomorphic to $U_p$ via $f_{1}$, similarly for $U_{q}$. Then your argument shows that the restriction of $f_{1} \times f_{2}$ to $U_{p}^{i} \times U_{q}^{j}$ is a diffeomorphism onto $U_{p} \times U_{q}$ for all pairs $(i,j)$. – t.b. Mar 2 '11 at 16:33
• you´re right. But my intention was only to show that a product of diffeomorphisms is a diffeomorphism. I suppose that he have already grasped how to solve the exercise. – Júlio César Mar 2 '11 at 19:00