On the wiki page about cobordism, it is stated that the cobordism of oriented 0-dimensional manifolds is $\mathbb Z$. That seem surprising since One can always draw a line between two points.
I tried to find a proof or reference with no success, could someone give me the proof or a reference (that is not hidden behind a no-preview google-books)?
Here is the proof.
First, an orientation on a point is simply a choice of $+1$ or $-1$ for the point. That is, I'm thinking of orientation as a map $sign:M\rightarrow \{\pm 1\}$. I will call $p\in M$ positive or negative depending on the value of $sign(p)$.
For a (likely disconnected) closed $0$-manifold $M$, we define $f:M\rightarrow \mathbb{Z}$ by $f(M) = \sum_{p\in M} sign(p)$. Of course closed means "compact and no boundary", so there are only a finite number of points. Thus, the sum is finite, so $f$ is well defined.
Proposition 1: Suppose $M$ and $N$ are closed $0$-manifolds which are oriented cobordant. Then $f(M) = f(N)$.
Proof: Let $W$ be a cobordism between $M$ and $N$, with orientation agreeing with that of $M$ and opposite that of $N$. Of course, $W$ is just a disjoint union of intervals.
Let $p\in M$. Then $p$ is the end point of an interval, which has another endpoint $q$. If $q\in M$ as well, notice that by removing this interval from $W$, we are left with a cobordism between $M\setminus\{p,q\}$ and $N$. Further, $p$ and $q$ must have opposite signs because they are both in $M$ and the orientation of $W$ agrees with the orientation on $M$, so $f(M) = f(M\setminus\{p,q\})$.
Repeating this for both $M$ and $N$, we may assume that $W$ has the property that for each interval, one end point is in $M$ and the other is in $N$. Given a point $p\in M$ and $q\in W$ connected by an interval in $W$, because $W$ has the opposite orientation of $N$, we must have $sign(p) = sign(q)$. Further, $W$ gives a bijection between $M$ and $N$. Hence $f(M) = \sum_{p \in M} sign(p) = \sum_{q\in N}sign(q) = f(N)$.
$\square$
Let $C = \{$diffeo types of $0$-manifolds$\}/\text{cobordism}$ Then Proposition 1 implies there is an induced map $g:C\rightarrow \mathbb{Z}$.
Proposition 2: The map $g$ is a group isomorphism.
Proof: To see it's a homomorphism, just pick representatives all of whose points have the same sign.
Surjectivity is also easy: given $z\in \mathbb{Z}$, let $M$ consists of $|z|$ many points all of the same sign (positive if $z >0$ and negative if $z< 0$)$.
Injectivity: Suppose $f([M]) = f([N])$. Again, choose representatives $M$ and $N$ with all points positive or all points negative (the sign is determined by the sign of $f(M)$). Because $f(M) = f(N)$, the cardinality of $M$ and $N$ must match. So, picking a background bijection from $M$ to $N$, we connect points in $M$ with points in $N$ using closed oriented intervals. Because all the points in $M$ and $N$ have the same sign, the orientation on one end is that of $M$ and the orientation on the other end is $-N$, so we have an oriented cobordism between $M$ and $N$. Thus, $[M] = [N]$.
$\square$
Just for kicks, reducing $f$ mod $2$ gives the following:
Proposition 3: The unoriented cobordism group is $\mathbb{Z}/2\mathbb{Z}$.
Everything works identically except injectivity. But the point is that since you no longer care about orientation, any pair of points in $M$ can be paired off and then connected by an interval.