Mathematicians love fixed point theorems. Brouwer fixed point theorem is one such result. It states that “for any continuous function **f** mapping a compact convex set into itself there is a point **a** such that **f(a) = a**.” We are interested in one such fixed point theorem.

Consider maps (continuous functions) from \( S^n \to S^n \) (n-sphere to n-sphere). We will come up with a ‘test’ which will say, when do a function **does not **have a fixed point.

The antipodal map **h **is defined as \( h(\bar{x}) = – \bar{x} \). The bar above **x** says that **x **may have more than one coordinate. For example a point on the 2-sphere has three coordinates (x, y, z). Hence **h(x, y, z) = (-x, -y, – z)**.

In the above picture, we have drawn a 2-sphere. **A **is any point on the sphere. **B **is its antipodal point. B is found by changing the sign of each coordinate of A. Geometrically B is found by drawing a line from A, through the origin. Wherever that segment hits the sphere again, is its antipodal point.

It is easy to check that **h **is a continuous map. Also, note that **h does not have a fixed point. **Since each point on the sphere is sent to some other point (the diametrically opposite point on the sphere), hence no point is left **fixed **by the map **h.**

There is a notion called **homotopy of maps **in advanced mathematics. The idea is this: suppose there are two maps **h **and **g **from X to Y. Can you continuously deform **h **to **g **? If yes, then the two maps are said to be homotopic. In more technical language, **h **and **g **are homotopic if, there is a map \( F: X \times [0,1] \to Y \) such that \( F(x, 0) = h(x), F(x, 1) = g(x) \) and F is continuous.

**(The word ‘map’ is used for ‘continuous function’).**

Now we will show that any map which does not have a fixed point, is homotopic to the antipodal map.

Consider \( F(x, t) = \dfrac{(1-t) f(x) – tx} { || (1-t)f(x) – tx || } \).

Here **f **is a map from n-sphere to n-sphere. Here **x **has n+1 coordinates.

Notice that F(x, 0) = f(x) (denominator is 1 as ||f(x)|| =1 , since f(x) is on the sphere).

Also F(x, 1) = – x, the antipodal map.

Clearly F is continous if the denominator is not zero. Let us find out, when the denominator is zero.

\( || (1-t) f(x) – tx || = 0 \implies (1-t) f(x) = tx \).

Taking the absolute value (norm) on both sides, we have \( |1-t| || f(x) || = |t| ||x|| \).

Since for **x **and **f(x) ** are on the surface of the n-sphere, hence their length (norm or absolute value) is 1. Thus we have \( (1-t) = t \).

This implies t = 1/2. But plugging in t = 1/2, in the equation (1-t) f(x) = tx, we have f(x) = x. Hence **f **, has a fixed point.

If **f **did not have a fixed point then, the denominator will never be zero, making it homotopic to the antipodal map.

There is a beautiful visualization of this above fact. Suppose **f **does not have a fixed point. Pick any point **a **on the n-sphere. We want to send it to **-a **(the antipodal point). The shortest path is along the great circle, but there are infinitely many ways to reaching **-a **from **a along a great circle **as **-a **is exactly opposite to **a**.

Hence to decide on, which direction to go, we will take the help of **f**. Since **f **does not have a fixed point, it will move **a **to some **a’. **Once **a **has moved a little in some direction (to **a’**), its path to **-a **becomes fixed (as it is nearer to **-a **along a certain great circle).

Hence any continuous function, from n-sphere to n-sphere, **that does not have a fixed point **is homotopic to (can be deformed continuously into) the antipodal map!