# Thread: Constructing homotopies of functions to the n-sphere

1. ## [SOLVED]Constructing homotopies of functions to the n-sphere

Here is problem statement:

Show that if X is any topological space, and if two continuous maps $\displaystyle f,g:X \to \mathbb{S}^n$ have the property that for all $\displaystyle x\in X$, $\displaystyle \| f(x)-g(x)\| < 2$, then f is (continuously) homotopic to g.

And here is a homotopy (in our case)
A function $\displaystyle F(x,t): X \times I \to \mathbb{S}^n$ such that $\displaystyle F(x,0)=f(x)$ and $\displaystyle F(x,1)=g(x)$.

I don't have any trouble coming up with a function that satisfies the "deforming" property...for instance, there is a smooth function (I won't prove it, but there is, see Milnor or any other diff. top. book)

$\displaystyle \phi (t)=0$ for $\displaystyle t\leq 1/3$
$\displaystyle \phi (t)=1$ for $\displaystyle t \geq 2/3$

And so our homotopy could simply be
$\displaystyle F(x,t)=\phi (t) g(x) + (1-\phi (t))f(x)$

For that matter, we could simply chose
$\displaystyle F(x,t)=\sin (t) g(x) + \cos (t) f(x)$

But the problem with both these functions is that they don't nessicarily satisfy $\displaystyle \| F(x,t)\| = 1$, which is required if we want the range to be the n-sphere. So does anyone have any suggestions on how to impose that condition, or any other approaches to this problem that might prove fruitful?

Thanks.

2. Notice that the condition $\displaystyle \|f(x)-g(x)\|<2$ is just another way of saying $\displaystyle g(x)\ne-f(x)$ (because the only way that $\displaystyle \|f(x)-g(x)\|$ could be equal to 2 would be if f(x) and g(x) are antipodal points on the sphere). So the line joining f(x) and g(x) does not pass through the origin. Therefore we can define $\displaystyle F(x,t) = \frac{tg(x) + (1-t)f(x)}{\|tg(x) + (1-t)f(x)\|}$. That's the required homotopy.

3. Looks good to me, thanks a ton!