Is the union of two smooth submanifolds in R^n smooth? What about their intersections? I am trying to reason a few ideas.


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*The unit circle in $\mathbb{R^2}$ is a smooth space, yet the parameterization self-intersects (given by $p(t) = (cos(t), sin(t)$). 


How are we able to reason that the unit circle is smooth in this case? If I understand this correctly , p(t) is both regular ($\nabla p(t) \neq 0)$  for all $t \in I \subseteq \mathbb{R}$, and $p(t)$ is not simple since it is not injective on the interior of I. (both cosine and sine are not injective)
Perhaps I don't understand why the question is claiming that the parameterization intersects. 


*Does the union of two smooth submanifolds in $\mathbb{R^n}$ need to be a smooth submanifold?


No, a connected submanifold is smooth if every point in the submanifold has a neighbourhood N on which the intersection of N with the submanifold is locally the graph of a $C^1$ function. 
It is not necessarily true that the union of two connected sets is connected, so it suffices to say that it is smooth if each of the connected submanifolds is smooth. However, this requires the submanifold to be in the same dimension for both submanifolds, otherwise we cannot guarantee that there would be no singularities. (I am not really sure how to expand on this!)
Does my logic seem reasonable? I am not sure if I am missing something key.


*What about the intersection of two smooth submanifolds in $\mathbb{R^n}$? Do they need to be  smooth submanifold?


No, as with the above logic.


*What if we require that both of the submanifolds being union/intersected have the same dimension?


In this case, we only require that each of the submanifolds to be smooth, for the union to be smooth. I also want to argue that the intersection would also be smooth, because then the points in both submanifolds that are locally the graph of a $C^1$ function are still locally the graph of a $C^1$ function under an intersection. 
I'd really appreciate to hear your thoughts on what I've said, and any other remarks you could add.
Thanks!
 A: For question 1, in your comment you mention two standard theorems that you can use to verify that a subset of $\mathbb{R}^n$ is a manifold. Each of them can be applied, but the manner in which they are applied is different.
If you want to use the parameterization theorem that you mention in your comment then you have to be careful to work locally. For instance, $p(t)$ is not injective on its entire domain $t \in \mathbb{R}$, but it is injective on various subdomains, although not on $[0,2\pi],$ because $p(0)=p(2\pi).$ For example, it is injective on $(0,\pi)$, which is enough to let you apply the parameterization theorem to conclude that the intersection of the unit circle with the half-plane $y>0$ is a manifold.
What about the rest of the circle? Well, rather than trying to force there to be a single parameterization on which $p(t)$ is injective, it is sufficient to cover the whole of $S^1$ by finitely many pieces on which the parameterization is injective. Then you will have expressed $S^1$ as a union of open sets on which it is a manifold, which is good enough to conclude that the whole of $S^1$ is a manifold.
Or, instead of using the parameterization, just use the pre-image theorem from your same comment, using the function $F(x,y)=x^2+y^2-1$.
For question 2, think of the union of the $x$-axis and the $y$-axis in $\mathbb{R}^2$. Each of them is smooth and connected, and their union is connected... but...
Questions 3 and 4 are somewhat deeper, and a complete answer to these questions requires a good understanding of transversality.
