In topology class, before we learned the homotopy-based proof of Brouwer's fixed point theorem, the professor mentioned the easy proof of the one-dimensional case: just draw the graph $f(x) = x$ in $[0,1]$ and use the intermediate value theorem.

Can this "easy" proof be generalized to a simpler proof of the theorem in $n$ dimensions? The idea is to draw the graph of $f(x_1,x_2,x_3,...,x_n)=(x_1,x_2,x_3,...,x_n)$ as a simplex in the space $[0,1]^{2n}$ and prove that any surface must cross the simplex, probably using a variation of the intermediate value theorem. I don't see this proof anywhere in Wikipedia's list, and it sounds simpler than the other proofs there - if it works, that is.

Unfortunately, I'm all out of four-dimensional paper, so I'm having trouble visualizing this for even the 2-D fixed point theorem. My question is: does this idea actually lead to a simple proof, or am I chasing down a blind alley?

  • $\begingroup$ Good question, but I'm 99% sure you're chasing down a blind alley. Perhaps someone else can explain why. $\endgroup$ – Potato Jan 2 '12 at 15:46
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    $\begingroup$ I don't think proving the analog of the intermediate value theorem for higher dimesnions willl be easier than simply proving the fixed point theorem. $\endgroup$ – Michael Greinecker Jan 2 '12 at 16:00
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    $\begingroup$ The problem is that there isn't such a thing as an "intermediate value" in two dimensions or higher. At least, not one that is useful for studying continuous functions. $\endgroup$ – Thomas Andrews Jan 2 '12 at 16:17
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    $\begingroup$ The intermediate value theorem is really a topological result; precisely, it arises because the only connected subsets of $\mathbb R$ are the intervals. In contrast, the space of connected sets in higher dimensions is much richer, i.e., have much less structure. In my limited understanding, this is why some rudimentary form of algebraic topology is needed to get to the fixed-point theorem. $\endgroup$ – Srivatsan Jan 2 '12 at 17:09

The thing is, homotopy theory is the two-dimensional version of the Intermediate Value Theorem. Here is the Intermediate Value Theorem:

Intermediate Value Theorem. Let $[a,b]$ be a closed interval. Let $f\colon [a,b] \to \mathbb{R}$ be a continuous function, and suppose that $f(a)$ and $f(b)$ lie on opposite sides of a point $q\in\mathbb{R}$. Then there exists a point $p\in [a,b]$ so that $f(p) = q$.

And here is the basic generalization of the Intermediate Value Theorem for functions in two dimensions:

Intermediate Value Theorem in Two Dimensions. Let $D$ be a closed disk, and let $\partial D$ denote its boundary circle. Let $f\colon D \to \mathbb{R}^2$ be a continuous function, and suppose that $f(\partial D)$ has nonzero winding number around a point $q\in\mathbb{R}^2$. Then there exists a point $p\in D\,$ so that $f(p) = q$.

A few notes:

  • In case you're not familiar with the concept, the winding number of a closed curve around a point $q$ in the plane is the number of times that the curve winds counterclockwise around the point. That is, the winding number is the element of $\pi_1(\mathbb{R}^2-\{q\})$ represented by the curve.

  • Here the requirement that $f(\partial D)$ winds around $q\,$ replaces the IVT hypothesis that the images of the endpoints of the interval lie on opposite sides of $q$. In both cases, the idea is that the image of the boundary somehow “surrounds” $q$.

There is even a generalization of the standard proof of the Intermediate Value Theorem to multiple dimensions.

Sketch of Proof of IVT: Suppose we bisect $[a,b]$ into two subintervals. Then one of these must also have the property that the images of its endpoints surround $q$. By repeatedly bisecting, we obtain a nested sequence of closed intervals whose intersection is a single point $p$. By continuity, $f$ must map $p$ to $q$.$\qquad\Box$

Sketch of Proof of IVT in Two Dimensions: Imagine the domain disk $D$ as a square, and imagine that we cut this square into four subsquares. Then the winding number of the image of the boundary of $D$ around $q\,$ is the sum of the winding numbers of the images of the boundaries of each of the four subsquares around $q\,$. In particular, if the winding number of $f(\partial D)$ around $q\,$ is nonzero, then the same must hold for at least one of the subsquares. Iterating this process, we obtain a nested sequence of closed squares, whose intersection must be a single point $p$, and it is easy to show that $f(p)$ must be $q$.$\qquad\Box$

Note that this proof contains an implicit algorithm for finding the root of the function $f\colon D \to \mathbb{R}^2$. This algorithm is a bit harder than the bisection algorithm in one dimension, since it involves computing some winding numbers.

Of course, none of this involves drawing the graph of $f\,$ in four dimensions, and for good reason: drawing the graph isn't even a particularly good way to prove the Intermediate Value Theorem in one dimension. Indeed, the only way to use graphs to prove the Intermediate Value Theorem is to invoke the much harder Jordan Curve Theorem. This method depends on the following result:

Path Intersection Theorem. Let $D$ be a closed disk, and let $\alpha\colon [0,1]\to D$ and $\beta\colon [0,1]\to D$ be continuous paths whose endpoints lie on $\partial D$. If the endpoints of $\beta\,$ lie on different components of $\partial D - \{\alpha(0),\alpha(1)\}$, then $\alpha$ and $\beta$ must intersect.

This theorem can be proven in a simple way using the Jordan Curve Theorem. It should be obvious how this can be used to prove the IVT.

Here is an analogous geometric theorem in four dimensions:

Disk Intersection Theorem. Let $D^4$ be a closed $4$-ball, and let $D^2$ be the closed unit disk. Let $\alpha\colon D^2 \to D^4$ and $\beta\colon D^2 \to D^4$ be continuous functions, and suppose that $\alpha(\partial D^2) \subset \partial D^4$ and $\beta(\partial D^2) \subset \partial D^4$. If the curves $\alpha(\partial D^2)$ and $\beta(\partial D^2)$ have nonzero linking number in the $3$-sphere $\partial D^4$, then the images of $\alpha$ and $\beta$ must intersect in $D^4$.

It seems to me that this theorem is harder than the Brouwer Fixed Point Theorem, but it does contain the essential geometry that must be used to prove the Brouwer Fixed Point Theorem (or the Two-Dimensional Intermediate Value Theorem) if you want to use graphs.


This sounds very similar to the proof of the theorem using Sperner's lemma. However, the geometric reasoning needed to prove the lemma for the $n$-dimensional simplex can hardly make it 'simple' (for my taste).

An outline of the proof can be found here.

This proof is indeed popular in game-theory courses, or basic (point-set) topology - since it doesn't use algebraic topology.

  • $\begingroup$ Of course. My bad. Well- that explains why I didn't find it in google... $\endgroup$ – yaakov Jan 2 '12 at 18:36
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    $\begingroup$ Little truly geometric reasoning is needed to prove the $n$-dimensional Sperner’s lemma; it’s an easy induction that just uses a parity argument. $\endgroup$ – Brian M. Scott Jan 2 '12 at 20:37
  • $\begingroup$ Link's broken. A fixed one's here: web.stanford.edu/~amwright/BFPT.pdf (I guess the author is at Stanford instead of U Chicago for some reason.) $\endgroup$ – Akiva Weinberger Nov 25 '15 at 14:19
  • $\begingroup$ @AkivaWeinberger: I have repaired the broken link with the one you provided. $\endgroup$ – Hans Oct 1 '17 at 7:33

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