# 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

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