The classification of surfaces

Let $H^{n} = \left\{ (x_{1}, \ldots , x_{n}) \in \mathbb {R}^{n} : x_{n} \geq 0\right\}$ and $\partial H^{n} = \left\{ (x_{1}, \ldots , x_{n}) \in \mathbb {R}^{n} : x_{n} = 0\right\}$. An $n$-manifold with boundary is a topological space $M$ such that:

1. there is an embedding of $M$ into $\mathbb {R}^{N}$ for some $N$;

2. given $x \in M$, there is a neighborhood $U$ of $x$ and a homeomorphism $h$ of $U$ onto an open set in $H^{n}$.

Show that if $h: U \to H^{n}$ is a homeomorphism onto an open set in $H^{n}$ and $h(x) \in \partial H^{n}$, then there is a smaller open set $V$ about $x$ and a homeomorphism $h': V \to H^{n}$.

In a manifold with boundary, a point $x$ with $h(x) \in \mathbb {R}^{n} \setminus \partial H^{n}$ (i.e.$h(x)$ in the interior of $H^{n}$) is called an interior point, and a point with $h(x) \in \partial H^{n}$ is called a boundary point. Assume the following two results (to be proved in Chapter 6 using homology theory): first, an open set in $\mathbb {R}^{n}$ is not homeomorphic to an open set in $\mathbb {R}^{m}$ for $m \neq n$; second, if $x \in \partial H^{n}$ and $U$ is an open set in $H^{n}$ about $x$, then $U$ is not homeomorphic to an open set in $\mathbb {R}^{n}$ (or even any $\mathbb {R}^{m}$).

Using these two results, show that the dimension of a manifold is well defined and that an interior point cannot be a boundary point as well, so the concepts of interior point and boundary point are well defined.

Suppose $U$ is an open set about $0 \in [0,\infty ) = H^{1}$ and $V$ is an open set in $\mathbb {R}$. Show that $U$ is not homeomorphic to $V$.

Show that the boundary of an $n$-manifold is either empty or an $(n-1)$-manifold.

Suppose $M, N$ are $n$-manifolds with boundary. Show that if $h: M \to N$ is a homeomorphism, then $h|_{\partial M}$ is a homeomorphism from $\partial M$ to $\partial N$.

Theorem (Invariance of domain): Suppose $U$ is an open subset of $\mathbb {R}^{n}$ and $f: U \to \mathbb {R}^{n}$ is $1$--$1$ and continuous. Then $f$ is an open map; that is, it maps open sets to open sets.

Apply this theorem to prove the following version for $n$-manifolds: if $M^{n}, N^{n}$ are $n$-manifolds and $f: M^{n} \to N^{n}$ is $1$--$1$ and continuous, then $f$ is an open map. Show that if we also assume that $M^{n}$ is compact, connected and $N^{n}$ is connected, then $f$ must be a homeomorphism.

Deduce the fact that the dimension of an $n$-manifold is well defined from the invariance of domain.

Prove that the invariance of domain holds when $n = 1$.

Show directly that $H^{2}_{+} = \left\{ (x_{1}, x_{2}) : x_{1} \geq 0, x_{2} \geq 0\right\}$ is homeomorphic to $H^{2}$.

Show that the product of an $m$-manifold $M$ and $n$-manifold $N$ is an $(m+n)$-manifold $M \times N$. Conclude that $T = S^{1} \times S^{1}$ (the torus) is a compact, connected surface.

Construct a homeomorphism between the quotient space $D_{1}^{1} \cup_{\mathrm{id}} D_{2}^{1}$ and the circle $S^{1}$, where $D_{i}^{1}$ denotes a copy of the interval $[-1,1]$ and the identification glues corresponding endpoints of the two copies together via the identity.

Construct a homeomorphism between the lower half of the disk and the upper half of the disk sending the boundary as indicated in Figure 2.11.

Two homeomorphic half disks.

Give a description of the Klein bottle $K$ as a quotient space of $\mathbb {R}^{2}$ analogous to the description of the torus $T$ as a quotient space of $\mathbb {R}^{2}$ given in Chapter 1. Use this description to show that $K$ is a surface.

Show that the Klein bottle $K$ is compact and connected.

Show by induction that if $D_{1}, \ldots , D_{k}$ are $k$ disjoint disks in the interior of a connected surface $M$, and $D_{1}', \ldots , D_{k}'$ is another collection of $k$ disjoint disks in the interior of $M$, then there is a homeomorphism $h: M \to M$ which is the identity on $\partial M$ sending $D_{i}$ to $D_{i}'$.

Recall that a homeomorphism $f: \mathbb {R} \to \mathbb {R}$ is called periodic of period $1$ if either $f(x+1) = f(x) + 1$ (for $f$ order preserving) or $f(x+1) = f(x) - 1$ (for $f$ order reversing). Given an order-preserving homeomorphism $f': [a, a+1] \to [b, b+1]$, define $f: \mathbb {R} \to \mathbb {R}$ by writing $x \in \mathbb {R}$ uniquely as $x = t + n$ with $n \in \mathbb {Z}$, $t \in [a, a+1)$, and setting $f(x) = f'(t) + n$. Show that the map $f$ defined this way is a periodic homeomorphism.

Given two disjoint arcs $I, I'$ in the circle $S^{1}$, choose intervals $J, J' \subset \mathbb {R}$ so that $p(J) = I$, $p(J') = I'$, where $p: \mathbb {R} \to S^{1}$, $p(t) = (\cos 2\pi t, \sin 2\pi t)$. The unique order-preserving affine linear homeomorphism from $J$ to $J'$ induces (via $p$) a standard'' homeomorphism from $I$ to $I'$. Show that this standard'' homeomorphism does not depend on the choice of $J, J'$.

Use the homeomorphism $h: D^{1} \to I$, $h(s) = \frac{1}{2}(s+1)$, and the fact that any order-preserving homeomorphism $f: I \to I$ is isotopic to the identity while any order-reversing one is isotopic to $r'(t) = -t$ (where $I = [-1,1]$), to prove the analogous statement for $D^{1}$:

Define an embedding of $D^{1}$ into $S^{1}$ to be orientation preserving (resp.reversing) if it is the composition of an order-preserving (reversing) embedding of $D^{1}$ into $\mathbb {R}$ and the map $p: \mathbb {R} \to S^{1}$, $p(t) = (\cos 2\pi t, \sin 2\pi t)$. A homeomorphism from $S^{1}$ to $S^{1}$ is orientation preserving if it comes from a periodic homeomorphism of $\mathbb {R}$ which is order preserving.

When a $1$-handle is attached to a single boundary circle of a handlebody, there are two standard ways of attaching it up to homeomorphism, pictured in Figure 2.17: on the left, the new boundary has two components in place of one; on the right, the new boundary still consists of one boundary circle, and there is a Möbius band embedded in the surface using a piece of the collar and the handle $h^{1}$.

Attaching a 1-handle to one boundary circle.

Find the Möbius band that is referred to above on the right surface in Figure 2.17.

Show that the result of attaching a 1-handle to the same boundary circle is never homeomorphic to what you get when it is attached to two different boundary circles.

Theorem: Let $H$ be a handlebody with a handle decomposition with $0$-handles and $1$-handles. Then: (a) $\partial H$ is homeomorphic to the disjoint union of a finite number of circles; (b) there is a neighborhood $N(\partial H)$ of $\partial H$ which is homeomorphic to $\partial H \times I$, where $\partial H \times \left\{ 0\right\} \subset \partial H \times I$ is sent to $\partial H \subset H$ via the standard identification $(x,0) \mapsto x$.

The proof outline is by induction on the number of $1$-handles: the case of $0$-handles alone is immediate, and for the inductive step one writes $H = H_{1} \cup h^{1}$, assumes (a) and (b) hold for $H_{1}$, and uses the existence of a collar where the $1$-handle is attached to determine the new boundary and its neighborhood, treating separately the case where the two attaching arcs of $h^1$ lie on the same boundary circle of $H_1$ and the case where they lie on two different boundary circles of $H_1$.

Fill in the details of this proof.

Show that the complement of the center circle in a Möbius band is connected but that the complement of the center circle in an annulus is separated.

Use the previous exercise to show that a Möbius band cannot be embedded in the plane.

An orientation for a handle is an orientation of its boundary circle, meaning a consistent ordering for any arc (a homeomorphic image of an interval in the circle): if two arcs intersect in an arc, the two orderings agree on the arc of intersection. In the Möbius band's handle decomposition with one $0$-handle and one $1$-handle, show that no matter how one orients each of the two handles, the $1$-handle is attached to the $0$-handle so that the orientation agrees on one of the two attaching arcs and disagrees on the other.

Use the definition of orientability of a handlebody (choosing an orientation for each handle so that the orientations disagree on all identifications of attaching arcs) to determine whether the handlebodies in Figure 2.25 are orientable or not.

Some handlebodies.

Show by induction on the number of handles that in a connected handlebody any two handles $h_{a}, h_{b}$ may be joined by a chain $h_{a} = h_{1}, h_{2}, \ldots , h_{k} = h_{b}$ with $h_{i}$ incident to $h_{i+1}$ (where two handles are called incident if they have boundary arcs which are identified).

Give an example of an oriented handlebody $H_{1}$ and a handle $h^{1}$ attached to $H_{1}$ so that $h^{1}$ cannot be oriented consistently with the chosen orientations of the handles in $H_{1}$, but $H_{1} \cup h^{1}$ is orientable.

Given two compact connected surfaces $M, N$ with boundary, choose an embedding $f$ of $\left\{ \pm 1\right\} \times D^{1}$ which sends $\left\{ -1\right\} \times D^{1}$ into an arc in $\partial M$ and sends $\left\{ 1\right\} \times D^{1}$ into an arc in $\partial N$. The boundary connected sum of $M$ and $N$, written $M \mathbin {\overline{\cup }} N$, is the surface $M \cup_{f} D^{1} \times D^{1} \cup_{f} N$ formed by attaching a $1$-handle $D^1 \times D^1$ joining $M$ and $N$ along these arcs.

If we choose different embeddings $i_{M}, i_{N}$ (in the analogous construction for the connected sum $M \# N$, where a disk is removed from each of $M, N$ and a cylinder $D^{1} \times S^{1}$ is glued in along the two resulting boundary circles) giving $i_{M}', i_{N}'$, then there are orientation-preserving homeomorphisms $h_{M}: M \to M$, $h_{N}: N \to N$ with $h_{M} i_{M} = i_{M}'$, $h_{N} i_{N} = i_{N}'$, induced by isotopies connecting $i_{M}, i_{M}'$ and $i_{N}, i_{N}'$. These induce a homeomorphism between the two ways of forming the connected sum, by sending $M \setminus \left\{ i_{M}(0)\right\}$ to $M \setminus \left\{ i_{M}'(0)\right\}$ via $h_{M}$ and $N \setminus \left\{ i_{N}(0)\right\}$ to $N \setminus \left\{ i_{N}'(0)\right\}$ via $h_{N}$.

Verify that this description of the homeomorphism in pieces does fit together to give a homeomorphism between the two ways of forming the connected sum with $i_{M}, i_{N}$ and $i_{M}', i_{N}'$; that is, check that it is consistent with the identifications being made to form the connected sums.

Construct a homeomorphism from $[-1,1] \times [-1,1]$ onto $[-1,1] \times [-1,0]$ which fixes pointwise $\left\{ -1\right\} \times [-1,0] \cup [-1,1] \times \left\{ -1\right\} \cup \left\{ 1\right\} \times [-1,0]$.

Show that $M \mathbin {\overline{\cup }} N \simeq N \mathbin {\overline{\cup }} M$, where $\mathbin {\overline{\cup }}$ denotes boundary connected sum of surfaces with boundary (Exercise 2.5.1).

Show that $(M \mathbin {\overline{\cup }} N) \mathbin {\overline{\cup }} Q \simeq M \mathbin {\overline{\cup }} (N \mathbin {\overline{\cup }} Q)$, where $\mathbin {\overline{\cup }}$ denotes boundary connected sum of surfaces with boundary (Exercise 2.5.1).

Show that $M \# N \simeq N \# M$, where $\#$ denotes connected sum of surfaces (possibly without boundary).

Show that $(M \# N) \# Q \simeq M \# (N \# Q)$.

Show that $M \mathbin {\overline{\cup }} D^{2} \simeq M$, where $\mathbin {\overline{\cup }}$ denotes boundary connected sum (Exercise 2.5.1) and $D^{2}$ is the closed disk.

If $N$ is a surface without boundary, and $N_{(p)}$ denotes $N$ with $p$ disjoint disks removed from its interior, show that $N_{(p)} \mathbin {\overline{\cup }} S_{(q)} \simeq N_{(p+q-1)}$, where $S$ is the $2$-sphere and $\mathbin {\overline{\cup }}$ denotes boundary connected sum (Exercise 2.5.1).

Show that $M \# S^{2} \simeq M$, where $S^{2}$ is the $2$-sphere.

If $K$ is the Klein bottle, use $K \simeq P \# P$ (where $P$ is the projective plane) to show that $K_{(1)} \simeq P_{(1)} \mathbin {\overline{\cup }} P_{(1)}$, where the subscript $(1)$ denotes removing one disk and $\mathbin {\overline{\cup }}$ is boundary connected sum (Exercise 2.5.1).

Let $B$ be the Möbius band. Construct a model of $B \# T$ (the connected sum of $B$ with the torus $T$) and show that it is homeomorphic to $B \# K$ (the connected sum of $B$ with the Klein bottle $K$).

Here $P^{(k)}_{(1)}$ denotes the boundary connected sum of $k$ copies of the projective plane $P$ with one disk removed, and similarly $T^{(k)}_{(1)}$ for $k$ copies of the torus $T$. Show that $P^{(k)}_{(1)} \mathbin {\overline{\cup }} T_{(1)} \simeq P^{(2k+1)}_{(1)}$. Conclude from this that $P^{(l)}_{(1)} \mathbin {\overline{\cup }} T^{(k)}_{(1)} \simeq P^{(2k+l)}_{(1)}$.

For each of the orientable surfaces (given by handle decompositions with a single $0$-handle and disjoint $1$-handles attached to its boundary) in Figure 2.46, use handle sliding to put it in standard form and identify the surface. Check your answer using the classification theorem (Theorem 2.6.2: a connected, orientable handlebody with $0$- and $1$-handles and Euler characteristic $\chi$ and $p$ boundary circles is homeomorphic to the standard model for $S_{(p)}$ or $T^{(g)}_{(p)}$, where $\chi = 2 - 2g - p$).

Surfaces for Exercise 2.6.3.

For each of the nonorientable surfaces in Figure 2.47, use handle sliding to put it in standard form and identify the surface. Check your answer using the classification theorem (Theorem 2.6.2: a connected, nonorientable handlebody with $0$- and $1$-handles and Euler characteristic $\chi$ and $p$ boundary circles is homeomorphic to the standard model for $P^{(h)}_{(p)}$, where $\chi = 2 - h - p$).

Surfaces for Exercise 2.6.4.

Consider the step in the proof of the classification theorem (Theorem 2.6.2) for the orientable case where a new torus pair (a pair of linked $1$-handles) is created by adding a new $1$-handle to a handlebody $H_{1}$ already in standard form $T^{(g)}_{(p)}$, giving $H = H_{1} \cup h^{1}$. Show that, if we regard the $0$-handle together with this new torus pair as giving us the torus-with-one-hole $T_{(1)}$, then we can remove this piece from the surface and replace it by a consistently oriented disk with the same boundary, to which the other handles (those of $H_{1}$ besides its $0$-handle) are attached. Considering this disk as the new $0$-handle and the other $1$-handles as attached to its boundary, forming a handlebody $H_{1}'$, show that $H \simeq H_{1}' \# T$. Then use induction to show that $H \simeq T^{(g+1)}_{(p)}$ when $H_{1}' \simeq T^{(g)}_{(p)}$.

Consider the step in the proof of the classification theorem (Theorem 2.6.2) for the orientable case where a new $1$-handle is attached to a single boundary circle of a handlebody $H_{1}$ (already in standard form) to form $H = H_{1} \cup h^{1}$. Show that if $H_{1}$ is homeomorphic to $T^{(g)}_{(p)}$, then we can add a $2$-handle to $H$ so that $H \cup h^{2} \simeq H_{1}$. Thus, removing this $2$-handle shows that $H \simeq T^{(g)}_{(p+1)}$.

Consider the step in the proof of the classification theorem (Theorem 2.6.2) for the nonorientable case where we attach a new $1$-handle $h^{1}$ to form $H$ from a handlebody $H_{1}$, which, by induction, may be assumed to be either $T^{(g)}_{(p)}$ or $P^{(h)}_{(p)}$ and already in standard form. Show that we can assume in either case that $h^1$ is attached away from the torus pairs or twisted handles of $H_1$, so it may be regarded as attached to $S_{(p)}$ (the disk with $p-1$ trivial handles). In the case where $H_{1}$ is orientable, $h^{1}$ will necessarily be attached to $S_{(p)}$ so that $S_{(p)} \cup h^{1}$ is nonorientable. Show that this is not necessarily the case when $H_{1}$ is nonorientable.

Show that if a $1$-handle $h^{1}$ is attached to a single boundary circle of $H_{1} \simeq S_{(p)}$ as a twisted handle to form $H$, then the union of the $0$-handle and $h^{1}$ forms a Möbius band $P_{(1)}$. Form a new handlebody $H_{2}$ by replacing $h^{0} \cup h^{1}$ by a disk with the same boundary and then attaching the other $1$-handles (those of $H_1$ besides its $0$-handle) to it. Show that $H_{2} \simeq S_{(p)}$ and $H \simeq P \# H_{2} \simeq P_{(p)}$.

Show that if a $1$-handle $h^{1}_{1}$ is attached to two boundary circles of $S_{(p)}$ in a twisted fashion to form $H$, and we choose another $1$-handle $h^{1}_{2}$ so that $h^{0} \cup h^{1}_{2}$ forms an annulus whose boundary consists of these two circles, then $H_{1} = h^{0} \cup h^{1}_{2} \cup h^{1}_{1}$ is a Klein bottle with a disk removed, $K_{(1)}$. Form a new handlebody $H_{2}$ by removing $h^{0} \cup h^{1}_{2} \cup h^{1}_{1}$ and replacing it with a disk (the new $0$-handle), then adding in all of the other handles of the original standard handle decomposition of $S_{(p)}$. Show that $H_{2} \simeq S_{(p-1)}$ and $H \simeq K \# S_{(p-1)}$.

Deduce the general induction step for the nonorientable case of the classification theorem (Theorem 2.6.2) from the constructions of the last three exercises (2.6.7--2.6.9), considering separately:

Classification theorem for surfaces: a compact connected surface (possibly with boundary) is homeomorphic to either a sphere with $p$ disks removed, $S_{(p)}$, a connected sum of $g$ tori with $p$ disks removed, $T^{(g)}_{(p)}$, or a connected sum of $h$ projective planes with $p$ disks removed, $P^{(h)}_{(p)}$. Orientability, the number $p$ of boundary components, and the Euler characteristic $\chi$ form a complete set of invariants. If the surface $M$ is orientable with $p$ boundary components and Euler characteristic $\chi$, then $M \simeq T^{(g)}_{(p)}$ with $g = (2 - \chi - p)/2$ (here $S_{(p)} = T^{(0)}_{(p)}$); if $M$ is nonorientable with $p$ boundary components and Euler characteristic $\chi$, then $M \simeq P^{(h)}_{(p)}$ with $h = 2 - \chi - p$.

Verify the formulas given above for $g, h$ in terms of $\chi , p$.

Consider the surface $S = h^{0} \cup h^{1}_{1} \cup h^{1}_{2} \cup h^{1}_{3} \cup h^{1}_{4}$, formed as a Möbius band with a $1$-handle attached (drawn as the right-hand surface of Figure 2.49: a rectangle $h^0$ with a bump pushed back in, giving a Möbius band, together with an extra handle that can be used to get another Möbius band). Show directly that this surface is $P^{(2)}_{(1)} \simeq P_{(1)} \mathbin {\overline{\cup }} P_{(1)}$ by finding two Möbius bands which are attached together by a $1$-handle.

Figure 2.50 shows a shaded surface with one boundary component, a knot embedded in $\mathbb {R}^{3}$ (pictured via a planar projection with over/undercrossings). It decomposes into three $0$-handles $h^{0}_{1}, h^{0}_{2}, h^{0}_{3}$ (disks cut off by line segments joining parts of the knot) and four $1$-handles $h^{1}_{1}, h^{1}_{2}, h^{1}_{3}, h^{1}_{4}$ (twisted rectangular strips in $\mathbb {R}^{3}$ joining these disks), giving Euler characteristic $\chi = 3 - 4 = -1$; orienting $h^{0}_{1}$ counterclockwise and $h^{0}_{2}, h^{0}_{3}$ clockwise, the handles $h^{1}_{1}, h^{1}_{2}, h^{1}_{3}$ can be oriented consistently with these, but $h^{1}_{4}$ cannot be oriented consistently with the rest, so the surface is nonorientable; it is homeomorphic to $P^{(2)}_{(1)}$ (a Möbius band boundary-summed with another Möbius band, obtained by noting that $h^{0}_{1} \cup h^{0}_{2} \cup h^{0}_{3} \cup h^{1}_{1} \cup h^{1}_{3}$ is five rectangles laid end on end, equivalent to a single rectangle; adding $h^{1}_{4}$ gives a Möbius band, and adding $h^{1}_{2}$ gives the Klein bottle with one hole).

Figure 2.51 shows another surface with the same knot as its boundary, built by an algorithm due to Seifert, again from three $0$-handles $h^{0}_{1}, h^{0}_{2}, h^{0}_{3}$ (with $h^{0}_{1}$ lying above the plane containing most of the knot, like the upper hemisphere of the sphere) and four $1$-handles $h^{1}_{1}, h^{1}_{2}, h^{1}_{3}, h^{1}_{4}$, connected by $h^{1}_{1}$ and $h^{1}_{3}$ so that, as before, these five handles fit together to give one rectangle.

Surface bounded by a knot.

Another surface bounded by the same knot, via Seifert’s algorithm.

By looking at Figure 2.51 as this rectangle with the two remaining $1$-handles $h^{1}_{2}, h^{1}_{4}$ attached, mimic the argument for Figure 2.50 above to show directly that the surface of Figure 2.51 is $T_{(1)}$ (it has one boundary circle, the knot, and Euler characteristic $-1$, so the handles $h^{1}_{2}, h^{1}_{4}$ can in fact be oriented consistently with the rest).

Determine which of the standard model surfaces the two surfaces in Figure 2.52 are homeomorphic to.

Surfaces to identify for Exercise 2.7.5.

For (b), the exterior region depicts a disk lying over the plane, which can be taken as a $0$-handle. The knot which they bound is called the trefoil knot. Use the classification theorem and then give simpler handle decompositions to exhibit these regions more clearly, in analogy to the worked example of Figure 2.50/2.51 (a shaded surface bounded by a knot, decomposed into $0$-handles given by disks cut off by line segments joining parts of the knot, and $1$-handles given by twisted rectangular strips joining these disks). For each surface, give a handle decomposition with the minimal number of handles.

Identify the following surfaces (in terms of the standard models $S_{(p)}$, $T^{(g)}_{(p)}$, $P^{(h)}_{(p)}$):

Identify the surfaces depicted in Figure 2.56.

Surfaces for Exercise 2.7.7: (a) a sphere with two holes and a 1-handle attached; (b) a disk with two holes and a 1-handle attached.

Identify the surface depicted in Figure 2.57.

Surface for Exercise 2.7.8.

Identify the surfaces in Figure 2.58, where the octagon's edges are identified in pairs as indicated by the matching letters and arrow directions.

Surfaces for Exercise 2.7.9.

Exercises 2.9.1--2.9.8 together outline an argument that any compact connected $1$-manifold is homeomorphic to $S^{1}$.

Verify that the circle $S^{1}$ is a compact connected $1$-manifold. Show that $S^{1}$ can be covered by two open sets $U, V$, each homeomorphic to $\mathbb {R}$, so that $U \cap V$ is homeomorphic to the union of two disjoint open rays.

Show that a compact $1$-manifold is not homeomorphic to an open set in $\mathbb {R}$.

Show that a proper open set in $\mathbb {R}$ may be decomposed (uniquely) as the union of a disjoint (countable) collection of intervals and rays.

Suppose $U_{i}, U_{j}$ are two open sets in a $1$-manifold $M$ with homeomorphisms $h_{i}: U_{i} \to \mathbb {R}$, $h_{j}: U_{j} \to \mathbb {R}$, and $h_{ji} = h_{j} h_{i}^{-1}: h_{i}(U_{i} \cap U_{j}) \to h_{j}(U_{i} \cap U_{j})$ is the homeomorphism between the two images of $U_{i} \cap U_{j}$. Show that if a finite interval $(a,b)$ is in the decomposition of Exercise 2.9.3 for $h_{i}(U_{i} \cap U_{j})$, then $h_{ji}(a,b) = \mathbb {R}$.

With $U_{i}, U_{j}, h_{i}, h_{j}$ as in Exercise 2.9.4, show that one of the following three cases must occur:

Show that in cases (a) and (b) of Exercise 2.9.5, $U_{i} \cup U_{j}$ is homeomorphic to $\mathbb {R}$.

Show that in case (c) of Exercise 2.9.5, $U_{i} \cup U_{j}$ is homeomorphic to $S^{1}$.

Use induction on the number of open sets in a covering of $M$ by open sets homeomorphic to $\mathbb {R}$, together with Exercises 2.9.6 and 2.9.7, to conclude that a compact connected $1$-manifold is homeomorphic to $S^{1}$.