There is a proof of Brauwer's fixed point theorem (for functions $f:\overline{D(0,1)}\subset\mathbb{R}^n\rightarrow \overline{D(0,1)}$), it appears in several sources including the wikipedia page. The idea is to construct a smooth function $r:\overline{D(0,1)}\rightarrow S^{n-1}$ such that $r|_{S^{n-1}}=Id$ : You assuming that $f$ has no fixed point, and let r(x) be the intersection point of the continuation of the line $[x,f(x)]$ with $S^{n-1}$ on the side of x. Showing it's smooth is not a big deal since $r(x)$ is of the form $r(x)=f(x)+\lambda(x-f(x))$ for $\lambda\geq 1$ and it clearly maps every point on $S^{n-1}$ to itself.
Then you take a volume $\omega$ form on $S^{n-1}$, and consider the pullback $r^*\omega$ and say that on one hand we have: $$0\underset{\text{volume form}}{<}\int_{S^{n-1}}\omega=\int_{S^{n-1}}{r|_{S^{n-1}}}^*\omega$$ and on the ohter hand: $$\int_{S^{n-1}}{r|_{S^{n-1}}}^*\omega=\int_{S^{n-1}}r^*\omega\underset{Stokes}{=}\int_\overline{D(0,1)}d(r^*\omega)=\int_\overline{D(0,1)}r^*(d\omega)=0$$ since $\omega$ is a maximal form on $S^{n-1}$, it's exterior derivative $d\omega$ is zero.
We have defined the pullback of a differential form using a smooth function $g:U\rightarrow \mathcal{M}$ for an open set $U\subset{\mathbb{R}^{n-1}}$ by $$g^*\omega(x)(v_1,\dots,v_{n-1})=\omega(g(x))(Dg(x)v_1,\dots,Dg(x)v_{n-1})$$ for any $x\in U$ and any vectors $v_1,\dots,v_{n-1}\in \mathbb{R}^{n-1}$
There is also a pullback with functions between manifolds using the local coordinates so I understand why $\int_{S^{n-1}}\omega=\int_{S^{n-1}}Id^*\omega=\int_{S^{n-1}}{r|_{S^{n-1}}}^*\omega$, but what I don't understand it is why can we say that $\int_{S^{n-1}}{r|_{S^{n-1}}}^*\omega=\int_{S^{n-1}}r^*\omega$ - how does the definition of a pullback by a smooth function extend to the boundry of the open set such that we can say state this equality.
More generally I start to wonder what do we mean when we pull-back differential forms from manifolds with boundry - is it the limit of the pull-backs from the interior, or do we pull them back using the induced atlas on the bounry? If it's the latter, how do we know we're getting a smooth differential form?
If $N\subset M$ is a (let's say) compact $k$-dimensional submanifold, let $\iota\colon N\to M$ be the inclusion map. Then when $\omega$ is a $k$-form on $M$, we have $\int_N \omega = \int_N \iota^*\omega$.
In the case you're asking about, $N=S^{n-1}$ and $M=D^n$. Since $r\circ\iota = \text{id}_{S^{n-1}}$ (by definition of a retraction), $$\int_{S^{n-1}}\omega = \int_{S^{n-1}}\text{id}^*\omega = \int_{S^{n-1}}(r\circ\iota)^*\omega = \int_{S^{n-1}}\iota^*r^*\omega=\int_{S^{n-1}}r^*\omega.$$