The first construction is the one the whole chapter is modelled on: take a closed interval and glue its two ends together. What comes out is the circle, and the proof is the recognition theorem applied to the map that wraps the interval once round.

Take a strip of paper and tape the ends together. Nothing was stretched, nothing was cut, and the result is a loop — but it is a loop for a reason that has to be stated, not for a reason that is obvious.

The instruction: keep every point of $[0,1]$ and declare the two ends to be the same point. Every other point keeps to itself, so the classes are the pairs $\{0,1\}$ and the singletons in between. The map $t\mapsto(\cos2\pi t,\sin2\pi t)$ glues exactly that pair and nothing else.

Write $S^1$ for the unit circle in the plane, with its topology from $\mathbb{R}^2$. The claim is that $[0,1]/\{0,1\}$, the interval with its two ends glued, is homeomorphic to . The map that proves it is , which wraps the interval once round.

• The wrapping map is continuous, onto, and identifies $0$ with $1$ and nothing else: two values of $t$ in $[0,1]$ give the same point exactly when they are the two ends.
• $[0,1]$ is compact by Heine–Borel and is Hausdorff as a subspace of the plane, so the recognition theorem applies.

The wrapping map is continuous because each coordinate is, and it is onto because every point of the circle is at some angle. Two points of the interval have the same image exactly when their angles differ by a whole turn, which inside this interval means they are the two ends. So the map glues exactly the pair the instruction glues. The interval is compact and the circle, being a subspace of the plane, is Hausdorff. The recognition theorem then identifies the glued interval with the circle.

Wrap the interval once round the circle.

Both coordinates are continuous, and every point of the circle is reached.

Angles agree only after a whole turn, which inside the interval means the two ends.

The interval is compact by Heine–Borel and the circle is a subspace of the plane.

The recognition theorem turns the map into a homeomorphism of the glued space.

The map that proves it sends $t$ to the point at angle $2\pi t$.

• $[0,1]$ with its ends glued is the circle.
• The proof is the wrapping map plus the recognition theorem.
• The gluing map is not open, and the same map on a half-open interval is not a homeomorphism.
• A function descends to the circle exactly when it agrees at the two ends.

Great work! You now know how to:

• glue the ends of an interval and name the result;
• give the map that proves it and check what it glues;
• see why the half-open interval fails;
• say which functions descend to the circle.