Surjection of map to n-Sphere Consider the map $f\colon S^n \times S^m \times [0,1] \rightarrow S^{n+m+1}$ defined by $f(p,q,t) = p \cos(\frac{\pi}{2}t) +q \sin(\frac{\pi}{2}t)$ where $p \in S^n$, $q \in S^m$ and $t \in[0,1]$. Show this map is surjective, where $n,m \in \mathbb{R}$
I have been able to show it when $n=m=0$ and for $n=0, m=1$, but I am having problems to generalize it.
Edit: Following Ted's suggestion, I think about $p,q$ as $p\in S^n\subset \Bbb R^{n+1}\times \{0\}\subset\Bbb R^{n+1}\times\Bbb R^{m+1} = \Bbb R^{n+m+2}$. So basically the map is similar to a linear combination of points
 A: Consider a point $(x,y) \in S^{n+m+1}$, where $x \in \mathbb R^{n+1}, y \in \mathbb R^{m+1}$. We are looking for a solution $(p,q,t) \in S^n \times S^m \times [0,1]$ of the two equations
$$\cos(\frac{\pi}{2} t) p = x, \sin  (\frac{\pi}{2} t) q = y .$$
Note that $\cos(\frac{\pi}{2} t), \sin(\frac{\pi}{2} t) \in [0,1]$ for $t \in[0,1]$. Taking the norm, we see that a necessary condition for $t$ is
$$(*) \quad \cos(\frac{\pi}{2} t) = \cos(\frac{\pi}{2} t) \lVert p \rVert = \lVert \cos(\frac{\pi}{2} t) p \rVert = \lVert x \rVert , \sin(\frac{\pi}{2} t) = \sin(\frac{\pi}{2} t) \lVert q \rVert = \lVert \sin(\frac{\pi}{2} t) q \rVert = \lVert y \rVert .$$
But $\lVert x \rVert^2 +  \lVert y \rVert^2 = \lVert (x,y) \rVert^2 = 1$. Thus $(\lVert x \rVert,\lVert y \rVert) \in S^1$. Hence there is a unique $\tau \in [0,2\pi)$ such that $(\lVert x \rVert,\lVert y \rVert) = (\cos \tau,\sin \tau)$. Since $\lVert x \rVert,\lVert y \rVert \ge 0$, we see that $\tau \in [0,\pi/2]$. Hence $t = \frac{2\tau}{\pi} \in [0,1]$ is the unique solution of $(*)$.
Moreover, if $x \ne 0$, we necessarily have $p = \frac{x}{\lVert x \rVert}$, and if $y \ne 0$, we necessarily have $q = \frac{y}{\lVert y \rVert}$.
By inserting we easily verify


*

*If $x,y \ne 0$, then $f(\frac{x}{\lVert x \rVert},\frac{y}{\lVert y \rVert},\frac{2\tau}{\pi}) = (x,y)$.

*If $x = 0$, then $\lVert y \rVert = 1$ and $\tau = \frac{\pi}{2}$. Let $p \in S^{n+1}$ be arbitrary. Then $f(p,y,1) = (0,y) = (x,y)$.

*If $y = 0$, then $\lVert x \rVert = 1$ and $\tau = 0$. Let $q \in S^{m+1}$ be arbitrary. Then $f(x,q,0) = (x,0) = (x,y)$.
