Show the disjoint union of 2 n-manifolds is an n-manifold If I have two n-dimensional manifolds X and Y, how do I show that the disjoint union of the two is also an n-dimensional manifold?
What I did so far
$X \sqcup Y = X \times \{0\} \cup Y\times \{1\} $
From the definition of a manifold there exists a homeomorphism that maps any local subset to an element of $\mathbb{R}^n$. Thus $$f_X: X \supset U_X \rightarrow \tilde{U}_X \subset \mathbb{R}^n$$ and similar for Y. If I define a function $f$ that is a composite of these two functions will I not end up with
$$f: X \sqcup Y \supset U_X \times \{0\} \cup U_Y \times \{1\} \rightarrow \tilde{U}_X  \times \{0\} \cup \tilde{U}_Y \times \{1\} \subset \mathbb{R}^{2n}$$?
Surely the image of $f$ must be of dimension $2n$. How can $f$ possibly be a homeomorphism if it maps from the disjoint union "product" space (which is essentially of dimension $2n$) into a space of dimension $n$.
Edit:
I think this question is more about the definition of a disjoint union. I always thought if $X = \{1,2\}$ and $Y = \{3,4\}$ then $X \sqcup Y = \{(1,0),(2,0),(3,1),(4,1)\}$ is essentially like a more primitive version of an outer product in a  vector space.
 A: Ok I think I understand where my mental block was. An outer product of the real line with the real line is effectively an uncountable number of disjoint unions of the real line with itself while the disjoint union can be mapped 1 to 1 back into the real line.
Here is very simple map from the space of $\mathbb{R}^n \sqcup \mathbb{R}^n \rightarrow \mathbb{R}^n$ since we can easily map $\mathbb{R}^n \rightarrow I^n = (-1,1)^n $ the n dimensional extended open unit interval. Since this is so we can obviously map the second $\mathbb{R}^n \rightarrow I^n = (-1,1)^n + a$ where $a$ is a constant of dimension $n$ with magnitude $>1$. Thus we have two disjoint open sets mapped to that are both a subset of $\mathbb{R}^n$ hence we complete the proof.
Notice this fails if we have an uncountable number of disjoint unions since there isn't enough "space" to fit all the $I^n$ without overlapping. This is a very non technical answer but I think it helps visualize the difference between the outer product and disjoint union at least.
