Prove that function from $S^n \times S^m \to 2S^{m+n+1}$ is immersion I would like to prove that the map $f: S^n \times S^m \to 2S^{m+n+1}: ((x_1,..,x_{n+1}), (y_1,...,y_{m+1})) \to (x_1,...,x_{n+1},y_1,...,y_{m+1})$ 
is an imersion. Here $2S^{m+n+1}$ is the $m+n+1$ dimensional sphere with radius $\sqrt2$. 
I know that I have to prove that the map $(f_\star)_p : T_p(S^n \times S^m) \to T_{f(p)}(2S^{m+n+1}) : [\gamma] \to [f \circ \gamma]$ is injective but I fail to do this. My initial idea was the following: 
Assume that $(f_\star)_p([\gamma_1]) = (f_\star)_p([\gamma_2])$. It holds that $[f \circ \gamma_1] = [f \circ \gamma_2]$ and thus $f \circ \gamma_1  \sim f \circ \gamma_2$. So there is a chart $(U,\phi: U \to \mathbb{R}^p)$ with $ U \subset \mathbb{R}^{n+m+2}$ open such that $(\phi \circ (f \circ \gamma_1))'(0) = (\phi \circ (f \circ \gamma_2))'(0)$. It is easy to see that $f:S^n \times S^m \to f(S^n \times S^m)$ is a homeomorphism and thus $\phi \circ f$ is a chart for $S^n \times S^m$. Thereby $\gamma_1 \sim \gamma_2$ and thus $[\gamma_1] = [\gamma_2]$. 
I don't think that this is true because we do not know if $\phi \circ f$ is globally good defined. Can someone help me? 
Thanks! 
 A: The map $h:\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+m+2}$
defined by $h((x_1,..,x_{n+1}),(y_1,..,y_{m+1}))=(x_1,..,x_{n+1},y_1,..,y_{m+1})$ is an immersion. Its restriction to the submanifold $S^n\times S^m$ of $\mathbb{R}^{n+1}\times\mathbb{R}^{m+1}$ is also an immersion.
A: Working in charts on spheres is (almost?) always painful. So, as another answer suggested, let's make it easier by looking at the map on Euclidean space. We can consider $F\colon\Bbb R^{n+1}\times\Bbb R^{m+1} \to \Bbb R^{n+m+2}$ given by $F(x,y) = (x,0)+(0,y) = (x,y)$, where we identify $\Bbb R^{n+m+2}$ with the product in a natural way. This is a linear map that truly is the identity map and is a diffeomophism. 
When we restrict to $M=S^n\times S^m$, $F|_M$ maps to $\{(x,y): \|x\|=\|y\|=1\}$, and so the image is contained in a sphere $N$ of radius $\sqrt2$, since $\|(x,y)\|^2 = \|x\|^2 + \|y\|^2$.
How do we check that $f\colon M\to N$ is in fact an immersion? Let $\pi\colon\Bbb R^{n+m+2}-\{0\}\to N$ be the obvious projection map given by $\pi(z) = \sqrt2 z/\|z\|$. Then we observe that $f = \pi\circ F|_M$. We want to check that $df_{(x,y)}$ is injective for any $(x,y)\in M$. Well, in view of our earlier discussion $dF_{(x,y)}$ is the identity map, and restricting the identity map to the subspace $T_{(x,y)}M$ is injective. What is going on with $d\pi_{(x,y)}$? It maps surjectively to $T_{(x,y)}N$ with kernel precisely the $1$-dimensional subspace of $\Bbb R^{n+m+2}$ spanned by $(x,y)$. So, if $v\in T_{(x,y)}M$ and $df_{(x,y)}(v) = d\pi_{(x,y)}dF_{(x,y)}v = d\pi_{(x,y)} v = 0$, this means that $v$ is some scalar multiple of $(x,y)$. But $(tx,ty)$ is tangent to $M$ if and only if $tx\in T_x S^n$ and $ty\in T_y S^m$, and this happens if and only if $t=0$, i.e., if $v=0$ to start with. Thus, $f$ is an immersion.
