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Let $h:S^{n-1}\to S^{n-1}$ be $C^{\infty}$ map. How to prove that a function $F: S^{n-1}\times[0,1]\to S^{n-1}$ given by

$$F(v,t)=(\cos{\pi t})v+(\sin{\pi t})h(v)$$

is proper $C^{\infty}$ map?

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up vote 1 down vote accepted

Assuming $F(v,t)\in S^{n-1}$ there is not much to show. Suppose $A$ is compact in $S^{n-1}$, so $A$ is closed. The map $F$ is obviously continuous, so $F^{-1}(A)$ is closed, and since $S^{n-1}\times [0,1]$ is compact $F^{-1}(A)$ is compact, so $F$ is proper.

The fact that $F$ is $C^{\infty}$ is obvious, too, it's the composition of $C^{\infty}$ maps.

One may ask why $F(v,t)\in S^{n-1}$, though. Is $\langle h(v),v \rangle = 0$?

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How do you go with proving that $F$ is continious and $C^{\infty}$? Is it with charts or there is something quicker? –  dmm Jul 29 '12 at 15:22
    
$h$ is $C^0$ (smooth), similarly $v\mapsto v$ and $t\mapsto \sin(t\pi), t\mapsto \cos(t\pi)$. $F$ is obtained by multiplying/adding these components (in $\mathbb{R}^{n}$). So if you know that $f+g:M\times N\mapsto \mathbb{R}^n$ is $C^0$ (smooth) if $f, g$ are $C^0$ (smooth), you know that $F$ is (this you should know from, e.g., class). Then if $S^{n-1}$ is viewed as submanifold of $R^n-\{0\}$ and you know that $F$ does not take on the value $0$ (cause it's values are of norm $1$), you have $F= P\circ F$, $P$ the radial projection to $S^{n-1}$. So F is smooth as composition of smooth maps. –  user20266 Jul 29 '12 at 15:43
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