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I am taking a class called calculus on manifolds, with pre requisites of Analysis in several real variables. I am having trouble with using the definition of a manifold which was taught in class as

Let $0

Suppose $M$ is a subspace of $\mathbb{R}^n$ such that for each $p\in M $, there is a set V containing $ p $, open in M and set U which is open in $\mathbb{R}^k$, and a continuous map $\alpha:U\rightarrow V$ such that; $\alpha$ is injective $\alpha^{-1}:V\rightarrow U$ is continuous $D\alpha(x)$ has rank k for each $x\in U$, $\alpha$ is of class $C^r, r\geq 1$

given some set described by a solution set, take $y^2=x^2+1$ for example.

How would I apply this to a solution set? I am finding it difficult to visualise this definition where there would be two co ordinate patches. I have tried taking $\alpha_i(x)=\pm\sqrt{(x^2+1)}$. This function is not injective. Does this mean that it is not a manifold? Or do I have to look at it locally? i.e. for a neighbourhood about every point of x? How would I go about showing this?

Does it suffice to show that for $U_1=\{x|x>0\}$ with the co ordinate patch $\alpha_1:U_1\rightarrow V$ is continuous for all x in U and is differentiable with class $C^{\infty}$ Then the same for the rest $x<0$ and the other coordinate patch? Thanks for any help.

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So your set is defined as

$$A=\{(x,y)\in\mathbb{R}^2\ |\ y^2=x^2+1\}$$

Define two functions $f_+, f_-:\mathbb{R}\to\mathbb{R}$, $f_+(x)=\sqrt{x^2+1}$ and $f_{-}(x)=-\sqrt{x^2+1}$.

You are right that neither $f_+$ nor $f_-$ is injective. However look at the definition, these can't be candidates for coordinate maps because the image is not a subset of $A$ (being itself a subset of $\mathbb{R}^2$). But you can construct coordinate maps out of them:

$$F_+:\mathbb{R}\to A$$ $$F_{+}(x)=(x, f_{+}(x))$$ $$F_-:\mathbb{R}\to A$$ $$F_{-}(x)=(x, f_{-}(x))$$

These are indeed injective. And they satisfy all necessary conditions (note that the image of each one of them is open in $A$).