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It's easy to embed $S^1\times S^2$ in $\mathbb{R}^5$, since $S^1\subset \mathbb{R}^2$ and $S^2\subset\mathbb{R}^3$, but $S^1\times S^2$ lives also in $\mathbb{R}^4$. How can we write the embedding map $S^1\times S^2\hookrightarrow\mathbb{R}^4$?

FreeFunctor
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    I don’t have a concrete embedding at hand, but I would try to adapt the usual embedding of the torus $\Bbb S^1\times \Bbb S^1 \subseteq \Bbb R^3$… – Jonas Linssen Feb 09 '22 at 10:44
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    See also https://math.stackexchange.com/questions/429614. – Qi Zhu Feb 09 '22 at 10:44
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    Consider any embedding $S^1\to \bf R^4$, let $C$ the image. Then the boundary of a tubular neigbourhood of C is $S^1\times S^2$, as the normal bundle is trivial (orientable). – Thomas Feb 09 '22 at 11:22

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Here is an explicit construction: $$(x,y,a,b,c)\mapsto ((2+x)a, (2+x)b, (2+x)c, y)$$ where $x^2+y^2=a^2+b^2+c^2=1$. Note that given

$(\alpha, \beta, \gamma, \delta)$ in the image, we can recover $x=\frac{\alpha^2+\beta^2+\gamma^2+\delta^2-5}{4}$, and hence $a,b,c$.

Just a user
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