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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.

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    $\begingroup$ The same question has already been asked last year: math.stackexchange.com/q/2767232. $\endgroup$ – Paul Frost Mar 22 at 16:14
  • $\begingroup$ @PaulFrost too bad it cannot be marked a duplicate... $\endgroup$ – Henno Brandsma Mar 22 at 21:55
  • $\begingroup$ Thanks. I couldn't see this post. $\endgroup$ – user625442 Mar 23 at 20:13
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    $\begingroup$ @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. $\endgroup$ – Paul Frost Mar 23 at 23:12

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