# h-cobordism-part 1

This series of posts are mainly based on the book ‘lectures on the h-cobordism theorem’ by John Milnor. And this post is an introduction to the h-cobordism, mainly the definitions and so forth.

H-cobordism is a branch of differential topology, which uses the cobordisms to define an equivalence between manifolds that is weaker than the equivalence of diffeomorphism.

Suppose that $M,M'$ two manifolds(here we assume always differentiable manifolds), if there is a compact manifold $N$ such that its boundary $Bd(N)=m\bigcup m'$, a disjoint union with $M$ diffeomorphic to $m$(with a diffeomorphism $F:m\rightarrow M$) and $M'$ diffeomorphic to $m'$(with a diffeomorphism $F':m'\rightarrow M'$). We call this quintuple $(N,m,m';F,F')$ a cobordism for $M,M'$. There are several things to say. $M,M'$ are not necessarily connected. If we assume them to be both connected, then $N$ has a boundary of two component. This is not a trivial condition. We can not always find such an $N$.

Now that we have defined objects(in a category to be clarified), then naturally we should consider morphisms between these objects. We define a morphism on the cobordisms for the pair $M,M'$ to be $f:(N,m,m';F,F')\rightarrow (N_1,m_1,m_1';F_1,F_1')$ where $f$ is a morphism of manifolds between $N$ and $N_1$ such that $f(m)\subset m_1, f(m')\subset m_1'$, and $F_1\circ f_m=F, F_1'\circ f'_{m'}=F'$. So in a natural way, we can define an equivalence relation on these triads for $M,M'$(of course we assume that the triads exist, which, though, is not always right). So, we thus defined a category for each pair $M,M'$.

There is another equivalence relation, though it is not so obvious. We say that $M$ is cobordant(a French word)  to $M'$ if there exists a cobordism $(N,m,m';F,F')$ for $M,M'$. Clearly, $M$ is cobordant to itself, and if $M$ is cobordant to $M'$(by one cobordism $(N,m,m';F,F')$), $M'$ is cobordant to $M''$(by a cobordism $(H,h,h';G,G')$), then we define $R=N\bigcup H/(F'(x)\equiv G(y), x\in m',y\in h)$ with the quotient differential structure. In fact we have to show that this differential structure exists, but here we will not prove it. Admitting this result, then $(R,m,h';F,G')$ give that $M$ is cobordant to $M''$. What is more, if $M$ is cobordant to $M'$, then $M'$ is cobordant to $M$, too. Thus an equivalence relation is established.

There is a finer structure in this equivalence relation. If $M$ is cobordant to $M'$ by a quintuple $c$, and $M'$ is cobordant to $M''$ by $c'$, then we define $cc'$ to be the quintuple described above. So, in this way, we define another category, the objects of which are the manifolds, and the morphisms are the cobordisms between manifolds(we have to show that for any manifold $M$, there is the identity morphism $id_M$. This is not hard, $M$ is cobordant to itself by $(M\times [0,1],M\times \{0\},M\times \{1\};id_M,id_M)$, which we set to be the identity morphism in this category).

In summary, we have defined two categories, the first is associated to two given manifolds, the objects being the cobordisms, the morphisms being the morphisms between cobordisms, the second category concerns with all the manifolds, the objects being the manifolds, the morphism being cobordisms.

Until now we have never talked about differential morphisms between these two manifolds $M,M'$. It is not easy to associate a function $f:M\rightarrow M'$ to the cobordisms on $(M,M')$. Note that in the definition of cobordism, we require that the two boundaries be diffeomorphic to these two manifolds, so perhaps we should also require that $f$ be a diffeomorphism. If it is so, then the quintuple $(M\times[0,1],M\times 0,M\times 1;id_M,f)$ defines a cobordism on $(M,M')$. And we denote this cobordism by $c_f$, induced by $f$. It is not hard to see that, if $f:M\rightarrow M',h:M'\rightarrow M''$, then we have that $c_{h\circ f}=c_fc_h$. So, in some sense, we give a (contravariant) functor from the category of manifolds with diffeomorphisms as morphisms to the category of manifolds with cobordisms as morphisms. But in general, the cobordisms on $M,M'$ are more numerous than the diffeomorphism between $M,M'$. So, in some sense, the cobordism is something softer than diffeomorphism.