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As determined here, there exist spaces covered by $S^n$ that can be immersed in $R^{n+1}$. However, the existence of an embedding is a stronger condition. I suspect it to be impossible.

Is there some sort of "trick"/easy proof?

What I've calculated so far:

For any space $M$ covered by the sphere, we have $H_{0}\cong H_{n}\cong\mathbb{Z}$.
All other homology groups are are torsion groups $T_m$, with $k*g=0,\forall g\in T_{m}$, where $k$ is the degree of the map from $S^n$ to $M$.

If $M$ is embeddable, then there exists $N$ such that $\partial N=M$.

With use of the Alexander duality theorem, we can determine the relationships between the homology groups of $M$ and those of $N$:

$T_{k}(M)\cong T_{k}(N)\times T_{n-k-1}(N)$, where $T_k$ represents the torsion-subgroup of $H_k$ .

Also, $H_k(N)=T_K(N)$ for $k>0$.

Additional : If $M$ admits an embedding, then it must admit an immersion. $M$ admits an immersion if and only if $M\times (0,1)$ is parallelizable.

user3257842
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1 Answers1

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There is exactly one 3-dimensional spherical space-form (apart from $S^3$) which embeds smoothly in $R^4$, namely, the quotient $S^3/Q_8$, where $Q_8$ is the quaternion group of order 8, see page 19 of "Embeddings of 3-manifolds in $S^4$ from the point of view of the 11-tetrahedron census" by Budney and Burton. I am sure, there are many examples in higher dimensions.

Moishe Kohan
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  • This is unexpected. Who knew that the boundary of the tubular neighborhood of the embedded projective plane is covered by $S^3$? Thank you. – user3257842 Sep 18 '17 at 07:35