# extend a continuous map on sphere to ball such that the image is out of a compact set

Let $$\phi: S^{k-1}\to E\setminus Q$$ be a continuous map, here

• $$E$$ is an infinite dimensional Hilbert space
• $$Q$$ is a compact set in $$E$$

I need extend $$\phi$$ to $$B^{k}$$ such that $$\phi: B^{k}\to E\setminus Q$$ is still continuous.

If $$Q := \{u\in E:\|u\|_E=1\}\setminus E_{k+1}~ \text{where}~ E_{k+1} ~\text{is any k+1-dimension subspace of} ~E$$ (not compact), that's Ok, how about other $$Q$$?

Extending $$\phi$$ to the ball is equivalent to proving that $$\phi:S^{k-1}\to H\setminus Q$$ is nulhomotopic.
To prove this, you can consider the map $$F:\phi(S^{k-1})\times Q\to E;(x,y)\to\frac{y-x}{|y-x|}$$, which has compact image, so there is some vector $$v$$ with $$|v|=1$$ outside the image of $$F$$. So you can homotope $$\phi(S^{k-1})$$ to $$\phi(S^{k-1})+kv$$ with $$k$$ as big as you want without intersecting $$Q$$ at any point. If you take $$k$$ big enough, $$\phi(S^{k-1})+kv$$ and $$Q$$ will be separated by a hyperplane (perpendicular to $$v$$, for example), so you can just linearly homotope $$\phi(S^{k-1})+kv$$ to a point in its side of the hyperplane without intersecting $$Q$$.