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.
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.