The only 1-manifolds are $\mathbb R$ and $S^1$

Question

I recall having heard somewhere that the only 1-manifolds (second countable, Hausdorff, connected spaces locally homeomorphic to $\mathbb R$) are $\mathbb R$ and $S^1$. Is this true? If so, is there a reasonably elementary proof?

Answer 1

There's a proof outlined in Problems 17-5 and 17-7 of John Lee's "Introduction to Smooth Manifolds" that uses a basic classification of integral curves of vector fields, specifically that a nonconstant maximally defined integral curve is either injective or periodic, which implies (after a small amount of work) that the image of any nonconstant integral curve is diffeomorphic to either $\mathbb{R}$ or $\mathbb{S}^1$. The problem is finished by showing that any 1-manifold is orientable, and thus admits a nonvanishing global vector field, of which you consider a maximally defined integral curve.

I don't think this is the same proof as given in Guillemin and Pollack or in Milnor, and for my money it's quite a bit simpler than both.

Answer 2

This is an extremely oft-duplicated post. However, I have not been able to find a clear proof here on MSE yet, and most answers just refer the reader to textbooks or outdated links.

Here's where I learned a clear proof from; except it wasn't always all that clear, so I'm writing it up to aid my own understanding and hopefully to help future visitors. This post is a little long because I've included helpful sketches to remind us what the hell we're even doing and have aimed for a conversational tone; I found the linked proof hard to read because of a lack of these two things.

Classification theorem: A connected Hausdorff second-countable space locally homeomorphic to either $\Bbb R^1$ or $\mathbb{H}^1=[0,\infty)$ at every point must be homeomorphic to $[0,1],[0,1),(0,1)$ or $S^1$.

If we consider $C^r$ manifolds (possibly with boundary, but always second countable and Hausdorff) then we can replace "homeomorphic" with "diffeomorphic".

If the second countable hypothesis is dropped, there are counterexamples such as the long line.

In particular, the only closed $1$-manifold is $S^1$ if one works without boundaries.

I will not prove the result for the $C^r$ or smooth cases i.e. check the homeomorphisms can be arranged to be diffeomorphisms, since that is a reasonably easy modification and done in the linked paper. I only seek to properly flesh out the details as trying to ensure all is rigorous caused me some headaches.

The proof starts with several lemmas about how distinct charts can overlap. We need to understand that so that we understand how to glue the intervals together. There is the uninteresting case where the local neighbourhoods contain each other, so we omit this. Let $M$ be a space satisfying the above hypotheses (second countability only matters at the very end and is not redundant in the compact case, but connectivity and Hausdorffness are crucial!) and let $U,V$ be nonempty open subsets of $M$, $\varphi:U\to\Bbb R,\,\psi:V\to\Bbb R$ embeddings, where the charts overlap; $U$ is not contained in $V$ and $V$ is not contained in $U$, but $U\cap V\neq\emptyset$.

With no loss of generality, $I:=\varphi(U)$ and $J:=\psi(V)$ may be taken to be bounded intervals, possibly with endpoint if $M$ has boundary. The results to follow mostly focus on handling boundary points; these results are mostly rigorous realisations of the following sketch:

The first three are possible configurations of $U$ and $V$ (which are to be interpreted as lines, but I have drawn thin dotted ellipses because I believe it's easier to visualise). In the first two, $U$ has a boundary point but we see the boundary is not in the interior of $V$; in the second, both $U$ and $V$ have a boundary; in the third, neither have boundary. The goal is to forbid the two cases which I've marked with a red cross: in the first, the issue is this; $U$ is open, and the boundary point of $U$ is in the interior of $V$ which is locally like $(0,1)$ and we know that something like $(0,1/2]$ is certainly not open in $(0,1)$! In the second, the problem is simply that $U$ contains $V$. In the third, as we desparately try to find a way to fit a boundary of $U$ in $V$, we encounter a new problem: the space is not connected! Why? Notice here that the intersection of $U$ and $V$ would just be that singleton boundary $\{p\}$ but this is closed - as $M$ is Hausdorff - and open, as $U,V$ are open, therefore $\{p\}$ is nonempty, clopen and not all of $M$, contradicting connectivity.

The real point is, if $U$ (or a component of $U$, if $U$ is not connected) has a boundary then in our setup that boundary point cannot be contained in $V$. Ok, enough talking. Notice $\varphi(U\cap V)$ is an open subset of $I$ and is thus a disjoint union of (relatively) open intervals, the disjoint union corresponding to the connected components.

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Let $K$ be a connected component of $\varphi(U\cap V)$ and $a$ an endpoint of $K$. Then $a\notin K$ and if $a\in I$ then $c=\lim_{t\to a\\t\in K}\alpha(t)\notin J$, where $\alpha:\varphi(U\cap V)\to\psi(U\cap V)$ is the transition homeo(diffeo)morphism $\alpha:=\psi\circ\varphi^{-1}$.

To intuit the second statement, consider the third example in my diagram. Let $a$ be the leftmost endpoint of $V$, which lands in $U$. $a\in I$ is true. Here all transitions are the identity, so the limit in question is just $a$ again; but if $a$ were in $J$ notice that would make a hard boundary of $V$ lying in the interior of $U$, which should be forbidden!

$\alpha|_K$ is a homeomorphism of bounded intervals and is thus strictly monotone and bounded, so the limit $c$ must exist (in $\Bbb R$). If $a$ is not an endpoint of $I$ then $a\notin K$ simply because $K$ is open in $I$ and we know what the open subintervals of $[0,1],[0,1)$ etc. look like. Suppose for the sake of contradiction that $a$ is an endpoint of $I$ and $a\in K$. $\alpha(a)$ is an endpoint of $\alpha(K)$, which is relatively open in $J$, so must be an endpoint of $J$. Let $b$ be the other endpoint of $K$. If $b\notin I$ then that means, as $K$ contains the other endpoint of $I$, that $I=[a,b)$ (or $(b,a]$) and $K=I$ but this could mean $U\cap V=U$ and we forbid this, so it follows $b\in I$. Similarly consider the other endpoint of $\alpha(K)$, $d:=\lim_{t\to b\\t\in K}\alpha(t)$; if $d\notin J$ then $J=[\alpha(a),d)$ (or $(d,\alpha(a)]$) and this equals $\alpha(K)$, implying $\psi(U\cap V)=\psi(V)$ i.e. $U\cap V=V$, another contradiction. Therefore $a\notin K$!

Suppose $a\in I$ but $c\in J$. As we know $a\notin K$, if $a\in\varphi(U\cap V)$ then $a$ lands in a connected component distinct from $K$ and is thus separated by a neighbourhood from $K$, contradicting the fact it's an (end, limit)point of $K$... so $\varphi^{-1}(a)$ exists but is not in $V$. Thus if $c\in\psi(U\cap V)$ then we can talk about $\alpha^{-1}(c)=\lim_{t\to a\\t\in K}\alpha^{-1}\alpha(t)=a$, contradicting $a\notin\varphi(U\cap V)$, so we see $\varphi^{-1}(a)$ and $\psi^{-1}(c)$ are in $U\setminus V$ and $V\setminus U$ respectively. In particular they are nonequal. As $M$ is Hausdorff, they are thus separable by neighbourhoods, but as both are $\lim_{t\to a\\t\in K}\varphi^{-1}(t)$ this is a contradiction; recall limits are not necessarily unique in non-Hausdorff spaces! Alternatively I could compute, in the instance $a\in I$ and $c\in J$ so that all is well-defined, $\psi^{-1}(c)=\lim_{t\to a\\t\in K}\psi^{-1}\psi\varphi^{-1}(t)=\lim_{t\to a\\t\in K}\varphi^{-1}(t)=\varphi^{-1}(a)$ using uniqueness of limits in a Hausdorff space, but this implies $\varphi^{-1}(a)\in V$ hence $a\in\varphi(U\cap V)$, forcing $a\in K$ by separatedness of the components, a contradiction.

Now we know $\varphi(U\cap V)$ is a union of relatively open intervals $K$ which do not contain their endpoints i.e. are truly open (in $\Bbb R$); supposing some $K$ has both endpoints in the interior of $I$ - equivalently, its endpoints are not endpoints of $I$ - we would find "$a\in I$" in both cases hence the corresponding "$c\notin J$" statement holds, meaning the endpoints of the interval $\alpha(K)\subseteq J$ are not in $J$ i.e. $\alpha(K)=J$, contradicting $V\not\subset U$. Hence $K$ must have at least one endpoint in common with $I$. Note a situation like $I=[a,x)$ and $K=(a,x)$ is very possible, but what we can deduce from this is that $\varphi(U\cap V)$ consists of one or two open intervals at most. Let's analyse these cases (doing this in detail is admittedly long and I apologise if that is frustrating, but hopefully it is clear!):

$(1)$ If $\varphi(U\cap V)$ consists of two intervals then $I,J$ are open and $U\cup V\cong S^1$. $(2)$ If $\varphi(U\cap V)$ consists of one interval and this interval has the same endpoints as $I$ then $I$ has exactly one boundary point and $U\cup V$ is homeomorphic to $[0,1]$ or $[0,1)$ according to whether or not $J$ has boundary. $(3)$ Otherwise, $U\cup V$ is homeomorphic to some interval $(0,1)$, $[0,1)$ or $[0,1]$ and $I,J$ contain at most one boundary point. $(4)$ If $U\cup V$ is homeomorphic to $[0,1]$ or $S^1$ then $M=U\cup V$.

To see $(1)$: let these be $K_1,K_2$ and let $a_i$ be the endpoint of $K_i$ which is not an endpoint of $I$ (so, $a_i$ lands in the interior) and let $b_i$ be the endpoint of $K_i$ which is an endpoint of $I$. Suppose $I$ is not open; intuitively this is bad because a situation like $I=\overset{=\varphi(U\setminus V)}{\overbrace{\{b_1\}\sqcup[a_1,a_2]}}\sqcup\overset{=\varphi(U\cap V)}{\overbrace{(b_1,a_1)\sqcup(a_2,b_2]}}$ would seem to contradict connectivity of $U$ and $V$; a path "in $V$" from $(b_1,a_1)$ to $(b_2,a_2)$ would have to cross the forbidden territory $\{b_1\}$ or $[a_1,a_2]$. In fact, we see such a path necessarily must "go around" $[a_1,a_2]$ which suggests why $U\cup V$ has to be $S^1$.

More precisely, if $I$ is not open (in $\Bbb R$) then without loss of generality we can say $b_1\in I$ and as $a_1\in I$ too, both instances of "$c$" from the first lemma must not be in $J$; but that implies the subinterval $\alpha(K_1)\subseteq J$ has both endpoints outside of $J$, forcing $\alpha(K_1)=J$ which is impossible. So, $I$ is open. If $J$ is not open then by a completely dual argument, with $K'_1=\alpha(K_1)$ and $K'_2:=\alpha(K_2)$ and replacing $I$ with $J$ in all the above (recall $\alpha$ is a homeomorphism $\varphi(U\cap V)\cong\psi(U\cap V)$) we realise the same kind of contradiction. So both $I,J$ are open intervals, or perhaps we should say arcs... without loss of generality $I=\overset{=\varphi(U\cap V)}{\overbrace{(b_1,a_1)\sqcup(a_2,b_2)}}\sqcup[a_1,a_2]$. Note that $a_1$ is possibly equal to $a_2$. Let $c_i=\lim_{t\to a_i\\t\in K_i}\alpha(t)$ and $d_i=\lim_{t\to b_i\\t\in K_i}\alpha(t)$. Suppose some $c_i$ is an interior point of $J$; then $\psi^{-1}(c_i)$ makes sense and equals $\lim_{t\to a_i\\t\in K_i}\varphi^{-1}(t)=\varphi^{-1}(a_i)$, contradicting $a_i\notin K_i$ as before. Hence! The $c_i$ are endpoints of $J$ and the $d_i:=\lim_{t\to b_i\\t\in K_i}\alpha(t)$ are interior points (possibly equal). $c_1\neq c_2$ because, as they'd both be endpoints - without loss of generality, the right endpoints - and this would imply, for some $x,y$, that $\alpha(K_1)=(x,c_1)$ and $\alpha(K_2)=(y,c_2)=(y,c_1)$ would intersect, contradicting injectivity of $\alpha$.

So $\alpha$'s monotone behaviour on $K_1,K_2$ must be the same, as it sends the $c_i$ to distinct endpoints, and without loss of generality we can say $\alpha$ strictly increases (on both intervals). If $d_1=d_2=d$ then $J=\overset{=\psi(U\cap V)}{\overbrace{(c_2,d)\sqcup(d,c_1)}}\sqcup\{d\}$ (and $a_1=a_2$) so if $z:=\psi^{-1}(d)$ then $U\cup V=U\cup\{z\}\cong(0,1)\cup\{z\}$ with the one-point compactification topology (check it! This is essentially because as you go arbitrarily close to $b_1,b_2$ within $U$, you go arbitrarily close to $z$) thus is homeomorphic to $S^1$. If $d_1\neq d_2$ then $d_2<d_1$ and: $$e^{i\theta}\mapsto\begin{cases}\varphi^{-1}(\lambda(\theta))&0<\theta<\pi\\\psi^{-1}(\mu(\theta))&\pi\le\theta\le2\pi\end{cases}$$Defines a homeomorphism $S^1\cong U\cup V$, if $\lambda:(0,\pi)\cong(b_1,b_2)=I=\varphi(U)$ and $\mu:[\pi,2\pi]\cong[d_2,d_1]=\psi(V\setminus U)$ are increasing homeomorphisms. Recall a continuous bijection from a compact space to a Hausdorff one is a homeomorphism!

$(2)$: If $I$ contains both its boundary points then $U$ is compact hence closed ($M$ is Hausdorff!) hence clopen but neither empty nor $M$, contradicting connectivity. If $I$ contains neither boundary point then $I=K$ by the presumption $K,I$ share endpoints, contradicting $U\not\subset V$. So $I$ is half open with a boundary point $a$. We know from the first lemma that $c=\lim_{t\to a\\t\in K}\alpha(t)\notin J$. With no loss of generality model $I=[a,b)$, $K=(a,b)$, $J=(c,d)$ or $J=(c,d]$ for some $b,d$. In other words, $J=\overset{=\psi(U\cap V)}{\overbrace{(c,e)}}\sqcup([e,d)\text{ or }[e,d])$ where $e=\lim_{t\to b^-}\alpha(t)$. The claim is then clear, by treating the cases $e=d$ and $e<d$ separately and gluing $(a,b)$ with $(c,e)$ along $\varphi,\psi$ as in case $(1)$.

$(3)$: Really this comes down to the three green ticks in my sketch. For the same reasons we know $I,J$ cannot contain both endpoints, and $I=[x,a)$ or $(x,a)$ where $K=(b,a)$ for some $b>x$ (wlog) and $J$ can be glued with $I$ along $(b,a)$ fairly happily, being perhaps careful with whether or not you choose increasing or decreasing parametrisations according to whether or not $\lim_{t\to b\\t\in K}\alpha(t)\notin J$ is the upper or lower endpoint of $J$. The reasoning is again similar to the explicit handling in $(1)$.

$(4)$: immediate from compact implies closed in the Hausdorff space $M$ and openness of $U\cup V$.

Now for the actual theorem. By second countability we may find a countable cover of $M$, $\mathscr{U}=\{U_n:n\in S\}$ for some initial segment $F$ of $\Bbb N$ with each $U_n$ equipped with some chart $\varphi:U_n\to\Bbb R$ having image a bounded interval. By Zorning on the poset of such covers ordered by refinement we find a minimal element and realise we may assume $\mathscr{U}$ already has the property $U_i\not\subset U_j$ when $i\neq j$, though possibly reducing the cardinality of $|S|$. Moreover we may order $\mathscr{U}$ in the following way: choose one boundary point of $M$, if one exists, and make $U_1$ contain that boundary point, and by connectivity we may assume $U_n\cap(U_1\cap U_2\cap\cdots\cap U_{n-1})\neq\emptyset$ for all $n$ - reorder like so by induction. How do we know we can reorder in such a way as to include every element of the cover? Well, if the inductive process, where we adjoin $U_k$ for the smallest permissible $k$ each time, doesn’t eventually give us all of the $U_\bullet$ we can repeat it again starting with $U_m$ for the smallest $m$ not included - note $U_m$ would have to be disjoint from all the included sets; and so on; if this happens then we have expressed $M$ as a (countable, but by more than one) union of disjoint opens, taking the union of the set of each such subordering, but this contradicts connectivity. So, we really can do this.

Inductively we see $U_1\cup U_2$ can have a chart $f_2$ identifying it with a bounded interval, $U_1\cup U_2\cup U_3$ also has some $f_3$, etc. until we terminate in $S^1$ or $[0,1]$ - in the cases $M$ is compact and $S$ is finite - or this just keeps going, if $S$ is infinite and $M$ is noncompact. Note each $U_n$ cannot contain $U_1\cup\cdots U_{n-1}$ and if the former is contained in the latter then we just keep the same chart, $f_n=f_{n-1}$ with the same image; else, the lemmae apply. Then we can take the pointwise limit of these chart functions $f_n$, this making sense since the chart for the union of the first $n$ $U_\bullet$ can be chosen compatible with the chart for the unions of the first $n-1$ $U_\bullet$ so the sequence $f_n(x)$ is eventually constant for any fixed $x$, and this witnesses $M\cong[0,1)$ or $M\cong(0,1)$ depending on whether or not $M$ had a boundary point (iff. $U_1$ had a boundary point); connectivity and noncompactness means the $U_{k\ge2}$ cannot have boundary points.