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When proving that Sperner's lemma implies the Brouwer fixed point theorem, we claim that since a n-dimensional ball is homomorphic to a n-simplex we can say that f, the continuous function, is continuous on the simplex and then we prove the Brouwer fixed point theorem for the simplex.

First, even if we find a fixed point for the n-simplex, how do we know that a continuous function on a n-dimensional ball would have a fixed point? What allows us to say that given that a continuous function on a simplex has a fixed point a continuous function on a n-dimensional ball would also have a fixed point?

Second, it is known that we can use Sperner's lemma to find fixed points, meaning that Sperner's lemma implies the Brouwer fixed point theorem in a somewhat constructive manner. I am assuming that this is done by searching for sequences of complete triangles and seeing where they converge. (They should converge to fixed points). Can we use this method to find fixed points not only on a simplex but also on a n-dimensional ball. If so, how would this be done? Thanks.

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  • $\begingroup$ The fixed point in the ball corresponds to the fixed point in the simplex under the homeomorphism. $\endgroup$ – Alex Provost Jul 31 '16 at 22:50
  • $\begingroup$ @AlexProvost I'm not sure what that means. Do you mean that if H is the homeomorphism from the n-simplex to the n-ball and f is continuous on the n-simplex and x is a fixed point (in the simplex) then H(x) is a fixed point in the n-ball? Does that mean that f(H(x)) = x? How can you prove that the homeomorphism 'preserves' fixed points? Thanks. $\endgroup$ – Yitzchak Shmalo Jul 31 '16 at 22:59
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    $\begingroup$ Any continuous $f:B^n \to B^n$ yields a continuous $H^{-1}fH:\Delta^n \to \Delta^n$. If $H^{-1}fH(x) = x$ then $f(H(x)) = H(x)$, i.e., $H(x)$ is a fixed point of $f$. $\endgroup$ – Alex Provost Jul 31 '16 at 23:32
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Let's write out the details. Suppose that $f : B \to B$ is a continuous function on the $n$-ball $B$. Let's let $S$ denote the $n$-simplex, so we also have a homoemorphism $$ H : S \to B $$

Then we get a map $$ g: S \to S: p \mapsto H(p) \mapsto f(H(p)) \mapsto $H^{-1}(f(H(P))) $$ i.e., $$g = H^{-1} \circ f \circ H.$$

Now suppose that we establish that $g$ has a fixed point $s \in S$, i.e., that $$ g(s) = s $$ for some particular $s$. Writing that out, we have \begin{align} g(s) &= s\\ H^{-1}(f(H(s))) & = s \\ f(H(s)) & = H(s) & \text{, by applying $H$ to both sides} \end{align} and we see that the point $b = H(s)$ is a fixed point for the original function $f$.

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