Fixing this proof of the simple connectedness of the $n$-sphere As we all know, the $n$-sphere ($n\ge 2$) is simply connected.
However, the way this was proven to me seemed somewhat complicated and I tried my hand at simplifying. My professor insists that my simplification is overlooking a key detail, but I don't understand what the failure is.
Can you explain it to me and if possible fix the mistake so that the proof is correct?

The proof in question:
Let $\gamma$ be a cycle of basis $p$ over the $n$-sphere.
Since the sphere is locally homeomorph to the space $R^n$ we can find a closed neighborhood $D$ at a point of the sphere which does not contain $p$ and it is homeomorph to a closed disk.

We can think of $\gamma$ as homotopic to the product of some paths either outside $D$ or inside $D$. But since $D$ is contractible we can find an homotopy of each of this paths inside $D$ to a path with the same end points but which only traverses the border of $D$ (which is path-connected since $n\ge 2$).

This induces a global homotopy of $\gamma$ to a path $\gamma'$ which evades the interior of $D$, and thus we can puncture the $n$-sphere in a point in the interior of $D$ to end in a space homeomorphic to $R^n$, which is again contractible and thus we can find an homotopy from $\gamma'$ to the constant path.
This shows that $\gamma$ is homotopic to a constant path, and thus the $n$-sphere is simply connected. $\square$
My professor points out that the step where the global homotopy is induced is not trivial, but I do not see how it fails.
 A: What you have is pretty much the argument you get by examining the proof of the van Kampen theorem which apply to this case.  It works out better if $D$ is an open set.  Let $x\in D$, let $U$ be an open disk containing $D$, and let $V=S^n-\{x\}$.  That way, $\{U,V\}$ is an open cover of $S^n$, with both $U$ and $V$ contractible (and $U\cap V$ is homotopy equivalent to $S^{n-1}$, but this does not matter for the proof --- all that matters is that it is path connected).
Since $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ are unions of open intervals and $I$ is compact, we can write $\gamma=\gamma_1*\gamma_2*\dots*\gamma_m$ for finitely many $\gamma_i$, with each $\gamma_i$ lying entirely within $U$ or $V$.  For a $\gamma_i$ lying entirely in $U$, as you note this path can be homotoped to lie within $U\cap V$, keeping the endpoints fixed (say by choosing such a path then doing a linear homotopy, viewing $U$ as being an open ball in $\mathbb{R}^n$).  Thus, every $\gamma$ is homotopic to a path lying within $V$, and $V$ is contractible.
