# Brouwer's Fixed Point Theorem: A proof with reduced homology

This post is about the proof I found very interesting during the Topology course I took this semester. It highlights the application of Reduced Homology, which is a modification of Homology theory in Algebraic Topology. We will use two results from Reduced Homology as black-boxes for the proof. Everywhere, we will assume $$\mathbb{Q}$$ is used as the coefficient of the Homology space.

Lemma 1 (Reduced Homology of spheres)
Given a $$d$$-sphere $$\mathbb{S}^d$$, then its reduced $$p$$-th Homology space is:

$$\square$$

Lemma 2 (Reduced Homology of balls)
Given a $$d$$-ball $$\mathbb{B}^d$$, then its reduced $$p$$-th Homology space is trivial, i.e. $$\tilde{H}_p(\mathbb{B}^d) = 0$$, for any $$d$$ and $$p$$.

$$\square$$

Equipped with these lemmas, we are ready to prove the special case of Brouwer’s Fixed Point Theorem, where we consider map from a ball to itself.

Brouwer’s Fixed Point Theorem
Given $$f: \mathbb{B}^{d+1} \to \mathbb{B}^{d+1}$$ continuous, then there exists $$x \in \mathbb{B}^{d+1}$$ such that $$f(x) = x$$.

Proof.    For contradiction, assume $$\forall x \in \mathbb{B}^{d+1}: f(x) \neq x$$. We construct a map $$r: \mathbb{B}^{d+1} \to \mathbb{S}^d$$, casting ray from the ball to its shell by extending the line segment between $$x$$ and $$f(x)$$. Observe that $$r(x)$$ is continuous because $$f(x)$$ is. Also, $$x \in \mathbb{S}^d \implies r(x) = x$$. Therefore we have the following commutative diagram. Above, $$i$$ is inclusion map, and $$id$$ is identity map. We then look of the Reduced Homology of the above, and this gives us the following commutative diagram. As the diagram commute, then $$\tilde{H}_d(\mathbb{S}^d) \xrightarrow{i^*} \tilde{H}_d(\mathbb{B}^{d+1}) \xrightarrow{r^*} \tilde{H}_d(\mathbb{S}^d)$$ should be identity map on $$\tilde{H}_d(\mathbb{S}^d)$$. By Lemma 2, $$\tilde{H}_d(\mathbb{B}^{d+1}) = 0$$. This implies $$\tilde{H}_d(\mathbb{S}^d) = 0$$. But this is a contradiction, as By Lemma 1, $$\tilde{H}_d(\mathbb{S}^d) = \mathbb{Q}$$. Therefore there must be a fixed point.

$$\square$$

## References

1. Hatcher, Allen. “Algebraic topology.” (2001).