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I need to show that $A = \{(x,s,t) \in \mathbb{R}^3 \; | \; x^4 + s^2 + t^2 = 1 \}$ is diffeomorphic to $B= \{(x, y,s,t) \in \mathbb{R}^4 \; | \; y=-x^2 \text{ and } x^2 + y^2 + y + s^2 + t^2 = 1 \}$.

A map $F: A \rightarrow \mathbb{R}^4$ defined as

$$ F(x,s,t) = (x, -x^2, s, t) $$ is smooth, injective, and onto $B$.

How do I show that both sets are diffeomorphic? The map $F$ is an immersion, but it's not clear to me how to use this to show that $A$ and $B$ are diffeomorphic.

nullbyte
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$A$ is compact, thus the map $f$ which is continuous and injective establishes a homeomorphism from $A$ to $B = f(A)$. You know that $A$ is submanifold of $\mathbb R^3$ and that $f$ is an injective immersion and a homeomorphism onto its image. This implies that $f$ is a diffeomorphism onto its image, i.e. an embedding of smooth manifolds.

Paul Frost
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  • Why do we need to establish homeomorphism? Isn’t an injective immersion between manifolds of same dimension a diffeomorphism? Or is it to show that both has same dimension? – Ivin Babu Feb 15 '22 at 16:10
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    @IvinBabu Injective immersions are in general no embeddings. See https://math.stackexchange.com/q/2067540. But you are right, if we have a bijective immersion between manifolds, then it is a diffeomorphism. But this requires a proof. See https://math.stackexchange.com/q/714579. And we have to show explicitly that $B$ is a manifold (this is easy). – Paul Frost Feb 15 '22 at 17:14