Extending a map from $S^n\to M^n$ to a nice map from $B^{n+1}\to M^n$ Let $S^n$ and $B^{n+1}$ be the unit sphere and unit ball in $\mathbb{R}^{n+1}$, and let $M^n$ be a contractible space of dimension $n$.
If necessary, assume that $M^n$ is a contractible simplicial $n$-complex.
Let $\sigma\colon S^n\to M^n$ be a continuous map.
Question
Does there exist a continuous map $\beta\colon B^{n+1}\to M^n$ such that for any $b\in B^{n+1}$, the point $b\in\mathbb{R}^{n+1}$ lies in the convex hull of the preimage $\sigma^{-1}(\beta(b))$?
Edit
The following stronger statement is false already if $n=1$.
There exists a continuous map $\beta\colon B^{n+1}\to M^n$ and an involution $f$ on $S^n$ with $\sigma=\sigma\circ f$, such that for any $b\in B^{n+1}$ there exists some $x\in S^n$ such that $\sigma(x)=\beta(b)$, and $b$ lies on the line segment between $x$ and $f(x)$ in $\mathbb{R}^{n+1}$.
The figure below illustrates a counter example to this stronger claim.
$Y$ is a star graph with three edges, and $\sigma$ is indicated by the dashed lines. None of the points interior to the white triangle lie between any pair of points that map to the same point on $Y$.

 A: The answer is no, essentially since higher homotopy groups of spheres are nontrivial.
For example, in the $n=3$ case let $M=B^3$ and let $\sigma\colon S^3 \to M$ be the Hopf map from $S^3$ to the boundary sphere $\partial M$ of $M$.  Viewing $S^3$ as the unit sphere in $\mathbb{C}^2$, the group $U(1)$ of unit complex numbers acts on $S^3$ by scalar multiplication, and the preimage of each point in $\partial M$ is a great circle in $S^3$ which is an orbit under the action of $U(1)$.
Now suppose $\beta\colon B^4\to M$ is any map satisfying the given conditions.  Note that $\sigma^{-1}(\beta(b))$ will be empty if $\beta(b)\notin \partial M$, so it must be the case that $\beta(b)\in \partial M$ for all $b\in B^4$, i.e. $\beta$ maps $B^4$ to $S^2=\partial M$.  Note that each $\sigma^{-1}(\beta(b))$ is one of the circles described above, and the associated convex hull is a disk through the origin.
Now, if $b$ is any point on the sphere $S^3\subset B^4$, then $b$ lies in the convex hull of $\sigma^{-1}(\beta(b))$ if and only if $b$ lies on the circle $\sigma^{-1}(\beta(b))$, which occurs if and only if $\beta(b)=\sigma(b)$.  Thus $\beta$ restricts to $\sigma$ on the boundary sphere $S^3$.  But this is impossible since $\beta$ maps $B^4$ into the $2$-sphere $\partial M$ and the Hopf map is not nullhomotopic.
