# Point-set topological proof of Brouwer's fixed point theorem

I have tried to understand the point-set topological proof of Brouwer's fixed point theorem presented in Cou11. But I couldn't clarify some parts. Here are the theorem and its proof.

Theorem: There is no retraction from the closed unit disk $$D^2$$ to its boundary the unit circle $$S^1$$.

Proof: We proceed by contradiction. Suppose $$r:D^2 \to S^1$$ is a retraction. Let $$a,b \in S^1$$. Note that $$S^1 \setminus \{a,b \}$$ is disconnected, as it is the union of two disjoint relative open sets. Thus, it suffices to show that it is the image under $$r$$ of a connected subset of $$D^2$$. Let $$A:=r^{-1}(a)$$. Now $$A \subset D^2$$ may be disconnected, but it certainly intersects $$S^1$$ at only one point, that is, at $$a$$, since $$r$$, restricted to $$S^1$$, is the identity. Similarly, $$B:=r^{-1}(b)$$ only intersects at $$b$$. The two sets $$A$$ and $$B$$ may intersect themselves, so excising them may disconnect $$D^2$$. However, because they each intersect $$S^1$$ at only one point, excising them leaves intact some subset of $$D^2$$ whose closure includes the entirety of $$S^1$$. Let this set be denoted by $$E$$. Now $$E$$ is connected, and it is open since $$A$$ and $$B$$ are closed. Thus, it is path connected.

Let $$a_0$$ and $$a_1$$ be endpoints of small arcs on $$S^1$$ centered at $$a$$. They are both contained in the closure of the path connected set $$E$$ discussed above, so there exists a continuous path $$γ$$ connecting them which does not intersect $$A$$ or $$B$$. But the union of this path with $$S^1 \setminus \{a,b \}$$ is connected —this can be seen by simply considering the union of $$γ$$ with some path from $$a_0$$ or $$a_1$$ to an arbitrary point on $$S^1 \setminus \{a,b \}$$. Finally, we note that $$r(S^1 \setminus \{a,b \} \cup \gamma) = S^1 \setminus \{a,b \}$$ because $$γ$$ avoided both $$A$$ and $$B$$. Thus, we have a contradiction.

Here are my questions:

1. "they each intersect $$S^1$$ at only one point, excising them leaves intact some subset of $$D^2$$ whose closure includes the entirety of $$S^1$$ ?"

2. "$$E$$ is connected, and it is open since $$A$$ and $$B$$ are closed ?"

3. "But the union of this path with $$S^1 \setminus \{a,b \}$$ is connected —this can be seen by simply considering the union of $$γ$$ with some path from $$a_0$$ or $$a_1$$ to an arbitrary point on $$S^1 \setminus \{a,b \}$$.?"

Any help will be appreciated.

• The same question has already been asked last year: math.stackexchange.com/q/2767232. – Paul Frost Mar 22 at 16:14
• @PaulFrost too bad it cannot be marked a duplicate... – Henno Brandsma Mar 22 at 21:55
• Thanks. I couldn't see this post. – user625442 Mar 23 at 20:13
• @M.GiovanniLucaretti When I answered to the quoted post, I didn't explicitly mention how poor the paper is. For example, "The two sets $A$ and $W$ may intersect themselves." But we have $A \cap W = f^{-1}(\{ \alpha\}) \cap f^{-1}(\{ \omega \}) = f^{-1}(\{ \alpha\} \cap \{ \omega \})) = f^{-1}(\emptyset) = \emptyset$ ;--) Perhaps there exists an elementary proof, but you will not find it in the quoted paper. – Paul Frost Mar 23 at 23:12