Show that, for any $n\geq1$ integer, $$S^1\times...\times S^1\subset\mathbb{R}^2\times...\times\mathbb{R}^2=\mathbb{R}^{2n}$$ can be embedded in $\mathbb{R}^{n+1}$. I first have to prove it without using explicit formulas, then by explicit formulas.
I find embeddings very difficult, I know the definition, but not how to work with it et cetera.. Who can help me?
An attempt:
Since a product of compact Hausdorff spaces is a compact Hausdorff space, let's call this space $X$. We then know, that an image of any closed subset of $X$ is compact, and thus under a continuous map it is compact - and if this continuous map is into $\mathbb R$, then it is closed, since again, since a compact subset of $R$ is closed. Therefore, using the embedding lemma, all we need to find is a continuous injection into $\mathbb R^{n+1}$, as such a map will seperate points and closed sets.
The idea behind the function we will use, is this: for $n=1$ we simply use the identity, for $n=2$ we use the torus, and for higher $n$ we generalize this - we keep "turning/rotating" the previous structure we've made, but we also add a new axis. By induction, we can then find a continuous injection from ${(S^1)}^n$ to $\mathbb R^{n+1}$ for any $n \in \mathbb N$.
In the following text, we're going to be considering the angle determining the point in $S^1$ as $\alpha$, instead of the coordinates of $S^1$ in $\mathbb R$.
The approach for finding the continuous injection, in more detail:
Say we have a continuous injection $f:{(S^1)}^n \to \mathbb R^{n+1}$, $f(\alpha_1,...,\alpha_n)=(f_1(\alpha_1,...,\alpha_n),f_2(\alpha_1,...,\alpha_n)...,f_{n+1}(\alpha_1,...,\alpha_n))$.
Define
$g(\alpha_1,...,\alpha_n,\alpha_{n+1})$ as
$([f_1(\alpha_1,...,\alpha_n)+A]\cos(\alpha_{n+1}), [f_1(\alpha_1,...,\alpha_n)+A]\sin(\alpha_{n+1}),f_2(\alpha_1,...,\alpha_n),...,f_{n+1}(\alpha_1,...,\alpha_n))$
Where $A$ is a number, such that $[f_1(\alpha_1,...,\alpha_n)+A]>B>0$ for all it's inputs. Such a number exists, because $f_1(\alpha_1,...,\alpha_n)$ is just a polynomial of $\sin$ and $\cos$ functions, so it is bounded.
$g$ is clearly continuous (again, it's a polynomial of $\sin$ and $\cos$), so all that's left, is showing that it is an injection: We take $\alpha=(\alpha_1,...,\alpha_{n+1})$ and $\beta=(\beta_1,...,\beta_{n+1})$, where $\alpha \neq \beta$, and for simplicity define $\alpha'=(\alpha_1,...,\alpha_{n})$ and $\beta'$ analogically.
If $\alpha_{n+1} \neq \beta_{n+1}$, notice that since $f_1(\alpha')+A=K$ and $f_1(\beta')+A=L$ are both positive, if we are to get $g(\alpha)=g(\beta)$, we need $(*)\cos(\alpha_{n+1}) (\frac{K}{L})=\cos(\beta_{n+1})$ and $(**)\sin(\alpha_{n+1}) (\frac{K}{L})=\sin(\beta_{n+1})$, so we then need $\alpha_{n+1}$ and $\beta_{n+1}$ to be in the same quadrant. But if they are in the first quadrant for example, then we observe that if $\sin(x) > \sin(y)$, we necessarily get that $\cos(x) < \cos(y)$ - and so it cannot happen that both $(*)$ and $(**)$ would be true at the same time, as $\alpha_{n+1} \neq \beta_{n+1}$. Therefore $g(\alpha)\neq g(\beta)$.
If $\alpha_{n+1} =\beta_{n+1}$, we have that $\alpha_i \neq \beta_i$ for some $i$ such that $1 \leq i \leq n$, and since we already presume that $f$ is an injection, this means that $g(\alpha)\neq g(\beta)$, as if $i \neq 1$ we get $g(\alpha)\neq g(\beta)$ directly from $f(\alpha')\neq f(\beta')$ , and if $i=1$, at least one of $\cos(\alpha_{n+1})$ or $\sin(\alpha_{n+1})$ is not zero, as $f_1(\alpha') \neq f_1(\beta')$, and so we also get that $g(\alpha)\neq g(\beta)$.
Therefore $g$ is an injection.
