One instance I am very fond of:

If $m,n$ are positive integers and $n$ is odd, then $S^m\times S^n$ has trivial tangent bundle.

Proof. Here the wise man is the tangent bundle of $S^n$. Think $T(S^m\times S^n)=TS^m\oplus TS^n$. Since $n$ is odd, $S^n$ has a nowhere vanishing vector field, so up to isomorphism $TS^n$ splits as $\mathbb{R}\oplus F$, where $\mathbb{R}$ denotes the trivial one-dimensional vector bundle.

Now the wise man $TS^n$ lends the trivial piece $\mathbb{R}$ to $TS^m$ and $TS^m\oplus\mathbb{R}$ becomes trivial, since (the pullback by $S^m\hookrightarrow\mathbb{R}^{m+1}$ of) $T\mathbb{R}^{m+1}$ equals $TS^m\oplus\mathbb{R}$, due to the presence of the outward unit normal.

So $TS^m\oplus\mathbb{R}=\mathbb{R}^{m+1}$ and now the wise man takes (a copy of) his $\mathbb{R}$ back to form again $TS^n$. Finally, he takes another $\mathbb{R}$ to trivialize himself in the same way.