I have read in this post Clarify: "$S^0$, $S^1$ and $S^3$ are the only spheres which are also groups" that there is a bijection between $S^1$ and the sphere $S^n$, $n \geq 1$ and this was confirmed in the answer!
What is an explicit bijection $f : S^1 \rightarrow S^n$ ?
Thanks!
There is no point seeking an explicit bijection. It will not have any special property. It will not be close to continuous.
$S^1$ and $S^n$ are in bijection because they have the same cardinality, and here is why:
If you remove a point from each of them, then you end up with $\Bbb R$ and $\Bbb R^n$ because of the stereographic projection.
Now, $\Bbb R$ clearly injects into $\Bbb R^n$, so by the Schröder–Bernstein theorem, we only need to inject $\Bbb R^n$ back into $\Bbb R$: given $(x_1, \cdots, x_n) \in \Bbb R^n$, let $0.a_{i1}a_{i2} \cdots$ be the binary expansion of $\frac1{1+\exp(x_i)}$ (this is the sigmoid function, an injection $\Bbb R \to (0, 1)$). Then say that the result is $0.a_{11} a_{21} \cdots a_{n1} a_{12} a_{22} \cdots a_{n2} \cdots$ as ternary. This different base is taken to avoid things like $0.012022222\ldots_3 = 0.121_3$. Then this gives an injection $\Bbb R^n \to \Bbb R$, so we are done.