# Mapping class group vs automorphism group in cobordism category

Let $3Cob$ be the category whose objects are closed surfaces and whose morphisms are diffeomorphism classes of cobordisms.

By sending a diffeomorphism $\phi$ of a surface $X$ to its associated cobordism cylinder $M_\phi$ we get a homomorphism $$M : \pi_0 (Diff(X)) \rightarrow Aut_{3Cob}(X)$$ which in general dimensions need not be injective or surjective (see eg. Dan Freed's lecture notes).

My understanding is that $M$ is indeed injective and surjective for $3Cob$. Is this true, and what is the easiest way to prove it?

As a follow-up question, what is known about the group $Aut_{nCob}(Y)$ for a closed $(n-1)$-manifold $Y$ for general $n$? Can it be expressed in terms of a more "traditional" automorphism group of $Y$? If not, does passing to the $\infty$-category setting give this question a nicer answer?

## 1 Answer

Injectivity is not hard to prove: (I'm quoting an email from Bruce.) Let f,g be pseudo-isotopic diffeomorphisms of a surface X. Then they are homotopic. Therefore they are isotopic. (see eg. Farb and Margulit, A primer on mapping class groups, pg 43).

Here is how you prove surjectivity; this amounts to proving that an invertible cobordism is a product (respecting all identifications). Let's call your surface F, so we have cobordisms A and B with $A\cup_F B = F \times I$ where $\partial A = F \cup F\times\{0\}$. (I am ignoring the actual identification maps you need to keep track of things in your category, but the argument would work as stated with more notation and care.) Then I claim that both A and B are product cobordisms. To see this, note that the projection $F \times I\to F \times \{0\}$ gives a map $F \to A \to F \times \{0\}$. The degree of this map is $1$, and so a standard covering space argument shows that the induced map $\pi_1(F) \to \pi_1(F \times \{0\})$ is surjective. By the Hopfian property of surface groups, it is an isomorphism. The same argument applies to the maps induced by including F into B.

Now van Kampen's theorem implies that the maps from F to A and B are isomorphisms on $\pi_1$, and so it follows that these are h-cobordisms. Since $F\times I$ embeds in $R^3$, it is irreducible (you can avoid this step by quoting the solution of the Poincaré conjecture, but that's overkill) and so work of Stallings (On fibering certain 3-manifolds. 1962 Topology of 3-manifolds and related topics (Proc. The Univ. of Georgia Institute, 1961) pp. 95–100) implies that they are both products. I believe that the proof of Stallings' theorem would actually give you a product respecting all identifications so that the resulting isotopy induces the given automorphism. You can also get this by Waldhausen's general theory of Haken manifolds (On irreducible 3-manifolds which are sufficiently large. Ann. of Math. (2) 87 1968 56–88), but Stallings' proof in this special case is easier.

Either proof of the product theorem would essentially proceed by finding a hierarchy for A and B that looks like you took a system of curves and arcs cutting F into a disk, and crossed with I. Since you know that $A\cup_F B = F \times I$, you can probably short-cut this argument by starting with such a hierarchy for $F \times I$, consisting of cylinders and disks. Then you would use standard 3-manifold arguments to say that these cylinders and disks can be assumed to hit F in circles and arcs in the same pattern as they do in $F \times \{0\}$. This (and irreducibility) would imply that the intersection of the cylinders and disks cuts each of A and B up into a ball, from which you get the product structure using Alexander's theorem.

• Thanks - accepting this answer, though it will take me a while to work through it. Any thoughts on general $n$? Jan 22 '14 at 17:11