## Factoring maps of handlebodies

Any map of finite graphs (1-dimensional CW-complexes) factors as a composition of

1. a finite sequence of folds;
2. an inclusion; and
3. a finite-to-one covering map.

There should be a corresponding result for handlebodies, which presumably should say that, after a homotopy, a continuous map of handlebodies factors as:

1. a compression (by which I mean a map of a handle into the complement of its interior);
2. an inclusion; and
3. a finite-to-one covering map.

Is my intuition correct, and does anyone have a reference? I'm specifically interested in how well-behaved the homotopy can be taken to be. For instance, can it be made to respect the boundary?

Notes

A fold is a map that identifies two edges with a common endpoint. Many folds don't change the homotopy type of a graph, and one would expect not to need these in the handlebody setting. The important folds are the ones that kill a loop. In handlebody terms, you can think of this as gluing in a two-handle, or as cutting a one-handle - hence my use of the word "compression". Is this word acceptable in this context?

The graph-theoretic result is due to Stallings.

By an inclusion of handlebodies, I mean that the new one should be obtained from the old by attaching 1-handles.

EDIT (prompted by Sam's comments below) I'm not quite sure what "respect the boundary" should mean, at this point. Suggestions welcome!

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Suppose that we are given a PL map from a handlebody $W$ to a handlebody $V$. Choose a spine for $W$. Homotope the map until the image is a regular neighborhood of the image of the spine. By general position, our map is now an embedding. Fix a pants decomposition of disks $D = (D_i)$, for $V$. Suppose that $P$ is a component of $V - D$ (so $P$ is a three-ball with three distinguished disks on its boundary). Consider a component $X$ of $f(W) \cap P$. This is essentially a knotted graph. Via a homotopy (keeping $X \cap D$ fixed) unknot $X$. If the rank of $X$ is positive then another homotopy produces small lollipops which we shall compress a bit later. If $X$ meets any disk $D_i \subset \partial P$ more than once then we may homotope a leg of $X$ through $D_i$. Let $Y$ be the resulting component of $f(W) \cap P$. (Note that this reduces $f(W) \cap D$.)

Homotoping in this fashion we eventually arrive at an embedding of $W$ so that every component in every three-ball of $V - D$ is either a tripod or an interval, possibly with lollipops attached. The feet of the tripod/interval lie in distinct disks in the boundary of the containing solid pants.

Now compress all of the lollipops to get $f'(W')$ (a new handlebody, because we compressed and a new map because we have to extend it over the two-handles we added).

EDIT: This reproduces, in our context, part of Stallings paper (eg sliding the leg is a fold, arriving at only tripods and intervals produces an immersion.)

Since $f'$ is an immersion, it follows from Stallings paper that $f'$ is $\pi_1$ injective and that $W'$ embeds into a finite cover of $V$.

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 Looks good! I want to use a homotopy to make X as big as possible in P, so the complement is either empty or a ball whose boundary contains one distinguished disc. If we do so then I think we get that, after a homotopy, the map factors as: glue on some 2-handles; glue on some 1-handles; and then a finite-sheeted covering. (Note that, in the question, I mentioned that by "inclusion", I meant gluing on 1-handles.) Does that sound right? – HW Nov 19 2009 at 16:27 (I mean that the complement of X in P should be a ball after X has been compressed.) – HW Nov 19 2009 at 16:31 This is right. After producing f' we can 1) isotope each components of $f'(W') \cap D_i$ to be homeomorphic to its containing disk, 2) apply isotopy extension to the map $f'$, and 3) fix the boundaries of the components $f'(W') \cap P$. – Sam Nead Nov 19 2009 at 17:54 Excellent! I think this is the best possible statement I could have hoped for. – HW Nov 19 2009 at 18:21