What is the problem with this "retraction" $D^2 \to S^1$?

Question

In Hatcher's Algebraic Topology, he proves Brouwer's fixed point theorem in dimension $2$.

He shows that if $f: D^2 \to D^2$ has no fixed points, then we can construct a retraction $r: D^2 \to S^1$. However, there cannot exist a retraction from $D^2 \to S^1$. So, $f$ must have a fixed point.

In the proof, the map looks like a perfectly fine map. But actually, it cannot be because there cannot exist such map.

However, I don't see what is wrong with this map. There cannot exist such a retraction, but in the proof, he constructs it. What is wrong with this $r$?

Answer 1

He can construct this $r$ precisely because we assume we never have $h(x)=x$.The construction is fine and continuous and well-defined under that assumption.

From this assumption comes a map we know cannot exist.

So the assumption was wrong.

It's a proof by contradiction. We don't construct a fixpoint, but derive a contradiction from the assumption we don't have any.

Answer 2

"In the proof, the map looks like a perfectly fine map. But actually, it cannot be because there cannot exist such map."

Which map looks perfectly fine? I'm assuming you mean the map $r$. But he doesn't actually construct a map $r$ -- he gives a recipe for constructing $r$, provided you have the right ingredients, and one of those is a map with no fixed points.

So: if you had such a map, you could build $r$. If I had a billion dollars, I could build a mansion in the middle of Central Park...but I don't, so it's moot.

Now you could try, given any map $f$, to build the associated $r$. Let's try that with $f(z) = iz$ (where I'm treating the plane as the complex plane). If $P = 1 + 0i$, then $f(P) = i$, and the arrow from $f(P)$ through $P$ hits the boundary of the circle...at $P$. Great!

What about if $P = 0.5 i$. THen $f(P) = -0.5 + 0i$, and the ray from $f(P)$ through $P$ hits the unit circle...uh...somehwere in the first quadrant.

This is looking pretty good: I pick a $P$, compute $f(P)$, and draw the ray from $f(P)$ through $P$ and see where it hits $S^1$.

OK. Now what if $P = 0 + 0i$. THen $f(P) = P$, and there is no "ray from $f(P)$ through $P$" or more precisely, there are infinitely many to choose from, and no choice makes the map we've been constructing actually continuous; to get continuity, we really do need $f(P) \ne P$ for every $P$.

Does that help at all?