From Opposite Points to a Zero on an Interval
On a circle, every point has a natural opposite point. The one-dimensional Borsuk-Ulam theorem says that a continuous real-valued function cannot distinguish every point from its opposite: at least one antipodal pair has the same function value. The useful idea is to compare the two values by subtraction. Along a semicircle, this difference varies continuously, and its values at the two endpoints have opposite signs unless one endpoint already gives equality.
We use the unit circle with its subspace topology inherited from the standard topology on \(\mathbb{R}^2\):
The point antipodal to \(z=(x,y)\in S^1\) is \(-z=(-x,-y)\). It lies on the circle because \((-x)^2+(-y)^2=x^2+y^2=1\). We parametrize the upper semicircle, including its endpoints, by
The identity \(\cos^2(\pi t)+\sin^2(\pi t)=1\) shows that \(\gamma(t)\in S^1\) for every \(t\in[0,1]\). Its endpoints are \(\gamma(0)=(1,0)\) and \(\gamma(1)=(-1,0)\), which are antipodal. The map \(\gamma\) is continuous, since its coordinate functions are continuous.
The function \(g\) is continuous: \(t\mapsto\gamma(t)\) and \(t\mapsto-\gamma(t)\) are continuous maps into \(S^1\); composing either with the continuous map \(f\) gives continuous real-valued functions; and the difference of two continuous real-valued functions is continuous. The endpoints have a crucial relation. Since \(\gamma(1)=-\gamma(0)\), we have
The One-Dimensional Borsuk-Ulam Theorem
The only real-variable result needed is the Intermediate Value Theorem: if \(h:[a,b]\to\mathbb{R}\) is continuous and \(r\) lies between \(h(a)\) and \(h(b)\), then \(h(c)=r\) for some \(c\in[a,b]\). In particular, if the endpoint values have opposite signs or either is zero, then \(h\) has a zero.
Proof. Let \(f:S^1\to\mathbb{R}\) be continuous, and form the continuous function \(g(t)=f(\gamma(t))-f(-\gamma(t))\) on \([0,1]\). If \(g(0)=0\), then \(f(\gamma(0))=f(-\gamma(0))\), and the theorem holds with \(z=\gamma(0)\). Otherwise \(g(0)\ne0\). The endpoint relation gives \(g(1)=-g(0)\), so \(g(0)\) and \(g(1)\) have opposite signs. The Intermediate Value Theorem, applied to the continuous function \(g\) and the value \(0\), gives \(t_0\in[0,1]\) with \(g(t_0)=0\). By the definition of \(g\), \(f(\gamma(t_0))=f(-\gamma(t_0))\). Taking \(z=\gamma(t_0)\in S^1\) proves the claim. \(\square\)
The argument covers both possible endpoint cases. If equality occurs at the starting endpoint, there is no need to invoke a sign change; otherwise the endpoint values are strictly opposite in sign and zero lies between them. No differentiability or formula for \(f\) is required. Continuity is what makes the Intermediate Value Theorem applicable.
Worked Applications
Worked Example: The First Coordinate
Let \(f:S^1\to\mathbb{R}\) be \(f(x,y)=x\). This function is continuous because it is the restriction of the continuous first-coordinate projection on \(\mathbb{R}^2\). Find an antipodal pair with equal values using the semicircle difference.
At \(\gamma(t)=(\cos(\pi t),\sin(\pi t))\), the function value is \(f(\gamma(t))=\cos(\pi t)\). At the antipodal point \(-\gamma(t)=(-\cos(\pi t),-\sin(\pi t))\), the value is \(f(-\gamma(t))=-\cos(\pi t)\). Therefore
At \(t=\tfrac12\), \(\pi t=\tfrac{\pi}{2}\), so \(g(\tfrac12)=2\cos(\tfrac{\pi}{2})=2\cdot0=0\). The corresponding points are \(\gamma(\tfrac12)=(\cos(\tfrac{\pi}{2}),\sin(\tfrac{\pi}{2}))=(0,1)\) and \(-\gamma(\tfrac12)=(0,-1)\). Substitution gives \(f(0,1)=0\) and \(f(0,-1)=0\), confirming equality.
Check: \(g(0)=2\cos(0)=2\), while \(g(1)=2\cos(\pi)=-2\). These values satisfy the required endpoint relation \(g(1)=-g(0)\), and the zero found at \(t=\tfrac12\) lies between them.
Worked Example: The Sum of the Coordinates
Let \(f:S^1\to\mathbb{R}\) be \(f(x,y)=x+y\). As a sum of coordinate functions, \(f\) is continuous. Along the semicircle,
and the antipodal point has both coordinates negated, so
Subtracting the second value from the first gives \(g(t)=2\cos(\pi t)+2\sin(\pi t)\). Take \(t=\tfrac34\), so \(\pi t=\tfrac{3\pi}{4}\). The exact trigonometric values are \(\cos(\tfrac{3\pi}{4})=-\tfrac{\sqrt{2}}{2}\) and \(\sin(\tfrac{3\pi}{4})=\tfrac{\sqrt{2}}{2}\). Hence
The corresponding points are \(\gamma(\tfrac34)=(-\tfrac{\sqrt{2}}2,\tfrac{\sqrt{2}}2)\) and \(-\gamma(\tfrac34)=(\tfrac{\sqrt{2}}2,-\tfrac{\sqrt{2}}2)\). At the first point, \(f=-\tfrac{\sqrt{2}}2+\tfrac{\sqrt{2}}2=0\); at the second, \(f=\tfrac{\sqrt{2}}2-\tfrac{\sqrt{2}}2=0\). Thus the pair verifies the conclusion directly.
Worked Example: A Function with Equality at the Endpoints
Let \(f:S^1\to\mathbb{R}\) be \(f(x,y)=x^2+2y\), again a continuous function because it is a polynomial in the coordinates. Its values at the paired points on the semicircle are
and
Therefore \(g(t)=4\sin(\pi t)\). At \(t=0\), \(g(0)=4\sin(0)=0\). The corresponding antipodal points are \(\gamma(0)=(1,0)\) and \(-\gamma(0)=(-1,0)\). Direct substitution gives \(f(1,0)=1^2+2\cdot0=1\) and \(f(-1,0)=(-1)^2+2\cdot0=1\).
In fact, this function agrees at an antipodal pair exactly when the pair lies on the horizontal axis. For an arbitrary \((x,y)\in S^1\), subtraction gives \(f(x,y)-f(-x,-y)=4y\). This difference is zero precisely when \(y=0\). The circle equation then gives \(x^2=1\), so \(x=1\) or \(x=-1\), producing exactly the pair \((1,0)\), \((-1,0)\). This direct check agrees with the semicircle calculation: \(4\sin(\pi t)=0\) at the endpoints \(t=0,1\), and is positive for \(0<t<1\).
Why the Sign Change Matters
The endpoint relation does not say that the antipodal difference has the same value at the two ends; it says the values are negatives of each other. That distinction is the mechanism of the proof. When the starting value is nonzero, the difference changes sign between the endpoints, so continuity forces a zero somewhere on the semicircle. A zero is exactly the desired equality of function values at opposite points.
Common pitfall. A mistaken calculation may treat \(g(1)\) as \(g(0)\), on the grounds that the parametrization goes around a circle. But this parametrization covers only a semicircle, and its endpoints are opposite points, not the same point. For the first-coordinate example, the actual calculations are \(g(0)=2\cos(0)=2\) and \(g(1)=2\cos(\pi)=-2\), so \(g(1)=-g(0)\), not \(g(1)=g(0)\). Using the wrong endpoint relation removes the sign-change argument. The correction is to evaluate the difference from its definition at each endpoint and use \(\gamma(1)=-\gamma(0)\).
The proof is a useful model for applying the Intermediate Value Theorem to a topological problem. A function on the circle may be complicated, but pairing each point with its opposite produces a scalar difference. The semicircle supplies an interval on which that difference is continuous, while its endpoints enforce the sign relation. The conclusion follows without choosing a special formula for the original function.
Check Your Understanding
Use the semicircle difference where appropriate. In each question, distinguish equality at an endpoint from a zero strictly between the endpoints.
- For \(f(x,y)=y\), calculate \(g(t)\), identify a zero, and verify the two function values at the resulting antipodal pair.
- For \(f(x,y)=x^2\), calculate the antipodal difference for an arbitrary \((x,y)\in S^1\). Which antipodal pairs have equal values?
- Suppose the antipodal difference for a continuous function has \(g(0)=0\). Why is it unnecessary to use the Intermediate Value Theorem?
- Explain why continuity of the antipodal difference is needed to conclude that opposite-sign endpoint values yield a zero.
- For \(f(x,y)=x-y\), compute the semicircle difference and find one parameter \(t\in[0,1]\) that gives equal values. Check the values at both points.