Why is Brouwer’s Fixed Point Theorem considered a result of algebraic topology? Brouwer’s Fixed Point Theorem and some generalisations can be proved using properties of Barycentric subdivisions and easy special cases of Sperner’s Lemma relating to them plus basic properties of metric spaces and the fact that every compact convex subset of $\mathbb{R}^n$ is homeomorphic to an $m$-simplex for $m\leq n$ which can be proved by projecting both onto the unit sphere and rotating. For example, see the Wikipedia entry for its generalisation The Kakutani Fixed Point Theorem which gives a sketch. Not so much as a homotopy, let alone any algebra. Why then is Brouwer’s Fixed Point Theorem considered part of algebraic topology when it is not about the objects of algebraic topology and it does not require the methods of algebraic topology?
 A: It is a result of algebraic topology because it can be easily proved by the machinery of algebraic topology (using homology groups or homotopy groups of spheres). However, there are alternative proofs based on simplicial triangulations, Sperner's lemma etc. By the way, the study of simplicial complexes was one of the starting points of algebraic topology (simplicial homology).
A similar example is the Fundamental Theorem of Algebra. There are many proofs using different approaches, for example via complex analysis, via algebraic topology (using e.g. the fundamental group), etc. In my opinion it is "a child of many parents" which doesn't make it wrong to call it, say, a result of complex analysis.
A: Sure, they can be proved combinatorially, but Brouwer invented degree to solve this problem. Brouwer certainly considered this to be a theorem of what we would now call algebraic topology. Is the fundamental theorem of algebra really a theorem of Lie theory because you can prove it by classifying compact Abelian Lie groups? Obviously not.
