Show that a function induces an embedding of the Klein bottle in $\mathbb{R^{5}}$ Show that the function $f :[0,2\pi]$×$[0,\pi]$ $\rightarrow$$\mathbb{R^{5}}$ defined by
\begin{equation}
 f(x,y)=(\cos x,\cos2y,\sin2y,\sin x\cos y,\sin x\sin y)
\end{equation} 
induces an embedding of the Klein bottle in $\mathbb R^5 $.So my attempt is to to use the next corollary:
Let $f:X \rightarrow Y $ be an onto map. If $X$ is compact and $Y$ is Hausdorff, then $f$ is an identification map.
So it's clear that $X$ is a compact space, and $Y= \operatorname{img}(X)\subseteq \mathbb{R^{5}}$. And this is Hausdorff. Then the corollary's hypotheses are satisfied.
It is an identification map. We know the Klein bottle can be obtained from $[0,2\pi]$×$[0,\pi]$ by identifying two opposite edges in the same orientation and the other two in the opposite orientation.
But I don't know how to identify the other points.
Thank you
 A: You identify $\{0\} \times [0,\pi]$ and $\{2\pi\} \times [0,\pi]$ with "the same orientation", i.e. you define $(0,y) \sim (2\pi,y)$. You identify $[0,2\pi] \times \{0\}$ and $[0,2\pi] \times \{\pi\}$ with "the opposite orientation", i.e. you define $(x,0) \sim (2\pi - x,\pi)$. This generates an equivalence relation $\sim$ on the closed rectangle $R = [0,2\pi] \times [0,\pi]$. The equivalence class $[x,y]$ of $(x,y) \in R$ is given by
$\bullet \quad [x,y] = \{ (x,y) \}$ if $(x,y) \in (0,2\pi) \times (0,\pi)$,
$\bullet \quad [x,y] = \{ (0,y), (2\pi,y) \}$ if $(x,y) \in \{0, 2\pi \} \times (0,\pi)$,
$\bullet \quad [x,y] = \{ (x,0), (2\pi - x,\pi) \}$ if $(x,y) \in (0,2\pi) \times \{0, \pi \}$,
$\bullet \quad [x,y] = \{ (0,0), (0, \pi), (2\pi, 0), (2\pi,\pi) \}$ if $(x,y) \in \{ 0,2\pi \} \times \{0, \pi \}$.
The quotient space $K = R/\sim$  is the Klein bottle.
We have $f(0,y) = (1,\cos 2y,\sin 2y,0,0) = f(2\pi,y)$ and $f(x,0) = (\cos x,1,0,\sin x, 0) = (\cos (2\pi - x),1, 0, \sin (2\pi -x)\cdot(-1),0) = f(2\pi - x,\pi)$.
In other words, $(x,y) \sim (x',y')$ implies $f(x,y) = f(x',y')$. Therefore $f$ induces a map $f' : K \to \mathbb R^5$. It remains to show that it is an embedding. Since $K$ is compact, it suffices to show that it is injective. This is equivalent to showing that $f(x,y) = f(x',y')$ implies $(x,y) \sim (x',y')$.
So let us consider the equation $f(x,y) = f(x',y')$. This gives us five equations, one for each coordinate. The second and third equation are $\cos 2y = \cos 2y'$, $\sin 2y = \sin 2y'$. Since $y,y' \in [0,\pi]$, this implies that either $y = y'$ or ($y = 0$ & $y' = \pi)$ or ($y = \pi$ & $y' = 0$).
Case 1. $y = y'$. Then the forth and fifth equation imply $\sin x = \sin x'$ because $\cos y$ and $\sin y$ cannot both be zero. Together with the first equation $\cos x = \cos x'$ we conclude (since $x,x' \in [0,2\pi]$) that either $x = x'$ or ($x = 0$ & $x' = 2\pi)$ or ($x = 2\pi$ & $x' = 0$). This means $(x,y) \sim (x',y')$.
Case 2.  ($y = 0$ & $y' = \pi)$ or ($y = \pi$ & $y' = 0$). Then the forth equation reduces to $\sin x = -\sin x' = \sin (-x') = \sin (2\pi -x)$. The first equation can be rewritten as $\cos x = \cos (-x') = \cos (2\pi -x)$. We conclude that either $x = 2\pi - x'$ or ($x = 0$ & $2\pi -x' = 2\pi$) or ($x = 2\pi$ & $2\pi -x' = 0$) because $x, 2\pi -x' \in [0,2\pi]$. The last two cases mean $x = x' = 0$ or $x = x' =2\pi$. Hence again $(x,y) \sim (x',y')$.
