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I'm trying to show that they're equivalent statements:

1) $1_{S^1}$ is not homotopic to a constant map.

2) $S^1$ is not a retract of $D^2$ ($D^2$ is the closed unit ball).

3) Every continuous map $f:D^2\to D^2$ has a fixed point.

1)$\to$2) I suppose the opposite, then there exists $r:D^2\to S^1$ continuous such that $r(x)=x$ for every $x\in S^1$. How can we show that $1_{S^1}$ is homotopic to a constant?

2)$\to$3) I suppose that there exists $f:D^2\to D^2$ such that $f(x)\neq x$ for every $x\in D^2$, but I don't know how to prove that $S^1$ would be a retract of $D$.

3)$\to$1) Of course I tried to suppose that $1_{S^1}$ is homotopic to a constant, and show a map without fixed points.

Any hint? Thanks.

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  • $\begingroup$ $1\to 2$, a retract gives you a homotopy equivalence, and $D^2$ is contractible. For $3\to 2$, take a fixed-point free map $f$ and define a map $g$ that takes $x$ to the point on the boundary where the line from $f(x)$ through $x$ meets it. That's a retract. I can see how to in the other direction, but can't think of a trick for $3\to 1$ directly at the moment... $\endgroup$ – Callus - Reinstate Monica Apr 24 '14 at 23:34
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For the equivalence $1\iff 2$:
A map $f:X\to Y$ is nullhomotopic iff there is a map $h:CX\to Y$ extending $f$. Now take $f=\text{id}_{S^1}$ and use that the cone $CS^1$ over $S^1$ is homeomorphic to the ball $D^2$.

For $2\iff 3$:
Suppose there is a fixed-point free $h$ on $D^2$. Then we can define a map $r:D^2→S^1$ by letting $r(x)$ be the point of $S^1$ where the ray starting at $h(x)$ and passing through $x$ leaves $D^2$. One can give a precise formula for $r(x)$, but it should be intuitively clear that $r$ is continuous. But this $r$ would be a retraction.
Conversely, a retraction $D^2\to S^1$ has $S^1$ as its set of fixed-points. Can you see how to compose it with a map on $S^1$ to make it fixed-point free ?

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