Cobordism of points 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)?
 A: 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.
