
# 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:

$\tilde{H}_p(\mathbb{S}^d) = \begin{cases} \mathbb{Q}, & \text{if } p = d \\ 0, & \text{otherwise} \enspace . \end{cases}$

$\square$

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).