Brouwer fixed-point theorem

Every continuous self-map of a closed disc has a fixed point.

If a continuous map $g:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has $\Vert g(x)-x\Vert \le K$, then it is onto. For a target $y$, choose $R\ge K+\Vert y\Vert$, $R>0$, and map the unit ball to itself by $f(x)=x-(g(Rx)-y)/R$. A fixed point gives $g(Rx)=y$.

Let $V$ be a finite-dimensional real inner-product space and let $F:V\to V$ be continuous. If $⟨F(x),x⟩<0$ on the sphere $\Vert x\Vert =R$, then $F$ has a zero in the open ball. Otherwise the radial projection of $F$ gives a map of the ball that contradicts the Brouwer fixed-point theorem.

If a polynomial equation can be rearranged as $z=F(z)$ with a continuous $F$ mapping a closed disk into itself, Brouwer's theorem guarantees a root in that disk.

If each component of a continuous map on a box has opposite weak signs on the corresponding pair of faces, the map has a zero.

Projecting a sign-corrected displacement map back to a square turns Brouwer's fixed point into a zero of the displacement, proving that paths joining opposite side pairs intersect.

Triangulate a triangle and label every vertex by $1$, $2$, or $3$, forbidding label $i$ on the side opposite vertex $i$. Then the number of small triangles carrying all three labels is odd, and in particular at least one such triangle exists. Counting edges whose endpoint labels are $1$ and $2$ proves the parity statement: boundary incidences are odd, while a small triangle has odd incidence exactly when it has all three labels.