Answer: Surjective Map on Interval Has Two Fixed Points

Key Idea / Intuition

The circle admits a rotation with no fixed point, so the answer to the first part is no — a simple rotation is a counterexample. For the second part, surjectivity is the key constraint: since $h$ maps onto all of $[0,1]$, the points $0$ and $1$ must each be "hit," and this forces the graph of $h$ to cross the diagonal $y = x$ at least twice. The argument is a clever two-application of IVT, exploiting the boundary behavior that surjectivity forces.

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Formal Proof / Solution

Part 1: The Circle Has No Fixed-Point Property

The rotation $g : S^1 \to S^1$ defined by $g(e^{i\theta}) = e^{i(\theta + \pi)}$ (rotation by $180°$) is continuous and has no fixed point, since $g(z) = -z \neq z$ for all $z \in S^1$.

So the fixed-point property fails for $S^1$.

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Part 2: Surjective $h : [0,1] \to [0,1]$ Has At Least Two Fixed Points

Setup. Define $\varphi(x) = h(x) - x$. We want to show $\varphi$ has at least two zeros in $[0,1]$.

Step 1: There is at least one fixed point.

Since $h$ maps $[0,1]$ to $[0,1]$, we have $h(0) \geq 0$ so $\varphi(0) \geq 0$, and $h(1) \leq 1$ so $\varphi(1) \leq 0$. By IVT, there exists $c \in [0,1]$ with $\varphi(c) = 0$, i.e., $h(c) = c$.

Step 2: Surjectivity forces $h$ to reach both endpoints, creating a second crossing.

Since $h$ is surjective, there exist $a, b \in [0,1]$ with $h(a) = 0$ and $h(b) = 1$.

• At $x = a$: $\varphi(a) = h(a) - a = 0 - a = -a \leq 0$, with equality iff $a = 0$.
• At $x = b$: $\varphi(b) = h(b) - b = 1 - b \geq 0$, with equality iff $b = 1$.

Step 3: Separate into cases.

Case A: $a = 0$ and $b = 1$.

Then $h(0) = 0$ and $h(1) = 1$, so both endpoints are already fixed points. That gives two fixed points immediately.

Case B: $a > 0$ (so $\varphi(a) < 0$) or $b < 1$ (so $\varphi(b) > 0$).

Without loss of generality suppose $a > 0$, so $\varphi(a) < 0$.

• We have $\varphi(0) = h(0) \geq 0$.
• $\varphi(a) < 0$.

By IVT, there exists $c_1 \in [0, a)$ with $\varphi(c_1) = 0$ — a fixed point strictly before $a$.

Now consider what happens after $a$: since $h$ is surjective, there exists $b$ with $h(b) = 1$, so $\varphi(b) = 1 - b \geq 0$. If $b > a$, then:

• $\varphi(a) < 0$ and $\varphi(b) \geq 0$, so by IVT there exists $c_2 \in (a, b]$ with $\varphi(c_2) = 0$.

Since $c_1 < a \leq c_2$, we have two distinct fixed points.

If instead $b \leq a$, we can apply a symmetric argument using $\varphi(1) = h(1) - 1 \leq 0$ and $\varphi(b) \geq 0$ with $b < 1$, again obtaining two zeros.

Conclusion. In every case, $h$ has at least two fixed points. $\blacksquare$

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Remark. The result is tight: $h(x) = x$ has infinitely many fixed points, and one can construct surjections with exactly two. The key is that surjectivity forces $h$ to "dip to $0$" and "reach $1$," creating two graph crossings with the diagonal.