Non-zero degree maps

Here's one of my favorite topics in low-dimensional topology.  

Let $M, N$ be closed, oriented, connected $n$-manifolds.  There's a lot of ways that $M$ and $N$ could be related: homeomorphic, cobordant, one could cover the other, etc.  The relation I'm interested in today is called ``domination''.  We say that $M$ dominates $N$ if there exists a non-zero degree map from $M$ to $N$.  (Recall that a map $f$ is degree $d$ if $f_*([M]) = d[N]$.)  For example, a covering map dominates with degree the number of sheets of the covering (up to sign).   

Here's a few cool facts about domination.

If $f: M \to N$ is degree $d$, then:

1) The image of $\pi_1(M)$ is finite index in $\pi_1(N)$ (and is surjective if the degree is 1).

2) $rank(H_i(M)) \geq rank(H_i(N))$ for all $i$.

3) If $M$, $N$ are hyperbolic 3-manifolds, then $vol(M) \geq d \cdot vol(N)$ with equality if and only if $f$ is homotopic to a cover.  (If you're familiar with the Gromov norm, the same inequality holds for the Gromov norm.)

As a general principle, if there is a non-zero degree map from $Y'$ to $Y$, then the invariants of $Y'$ should be bigger/more complicated than $Y$.  

I thought it might be fun to prove the first two.  [The third one goes by the Gromov norm which will have to be a post for another time.]

Proof of 1): We can factor $f$ through the cover of $N$ corresponding to $f_*(\pi_1(M))$.  If it was infinite index, $f$ would factor through a non-compact manifold, i.e. one with $H_n = 0$, and so $f_*([M]) = 0$.  Easy peasy.  You can check that the same factor-through-the-cover argument shows that $f_*(\pi_1(M)) = \pi_1(N)$, since otherwise the degree 1 map would factor through a map with higher degree (namely the associated cover).

Proof of 2:) Let's show that $f^* : H^i(N;\mathbb{R}) \to H^i(M; \mathbb{R})$ is injective.  Let $\alpha \neq 0 \in H^i(N;\mathbb{R})$.  Now put your Poincare duality hat on.  That means we can find $\beta \in H^{n-i}(N;\mathbb{R})$ such that $\alpha \cup \beta = [N]^*$, where $[N]^*$ is the cohomology class that evaluates to 1 on $[N]$.  Then, $$\langle f^*(\alpha \cup \beta), [M] \rangle = \langle \alpha \cup \beta, f_*([M]) \rangle = \langle \alpha \cup \beta, d [N] \rangle = d \neq 0.$$  Since $$f^*(\alpha) \cup f^*(\beta) = f^*(\alpha \cup \beta) \neq 0,$$ we get $f^*(\alpha) \neq 0$.  Yey!         

It turns out that in low dimensions, non-zero degree maps have a pretty nice structure.  Then here are some useful characterizations:

a) (Edmonds) Every non-zero degree map between surfaces is homotopic to the composition of a pinch map followed by a branched cover.  Here, a pinch map on a surface $S$ consists of writing $S = S_1 \cup_C S_2$ where $C$ is a separating curve, and quotienting $S$ to $S/S_2$.  

b) (Haken, Kneser, Waldhausen) If $f: M^3 \to N^3$ is degree 1, then $f$ is homotopic to a "pinch".  In other words, $N$ has a Heegaard splitting $N = H_1 \cup H_2$ and $M = U_1 \cup H_2$, and $h$ is the identity on the $H_2$ half and pinches the $U_1$ part onto the handlebody.  (More precisely, if the Heegaard splitting of $N$ is genus $g$, then there's a pair of curves in the boundary of $U_1$ that bound surfaces in $U_1$; map the surfaces to disks and the rest of the interior of $U_1$ to a ball.  This defines a map from $U_1$ to a handlebody.)  

c) (Edmonds) If $f: M^3 \to N^3$ is degree at least 3 and the induced map on $\pi_1$ is surjective, then $f$ is homotopic to a branched cover.   

d) (Gadgil) There exists a degree 1 map $f: M^3 \to N^3$ if and only if $M$ embeds in $N \times I \#_k \mathbb{C}P^2 \#_m \overline{\mathbb{C}P^2}$ for some $k, n$ if and only if $M$ is obtained by surgery on a nullhomotopic link in $N$.   

e) (Yi Liu) Given a fixed degree $d$ and a fixed three-manifold $M$, there are at most finitely different $N$ such that there exists a degree $d$ map from $M$ to $N$.

Let me give an alternative four-dimensional perspective on degree 1 maps due to Hee-Jung Kim and Danny Ruberman.  

Claim:  If $Y'$ is a 3-manifold obtained by surgery on a nullhomotopic link in another 3-manifold $Y$,  then there exists a degree one map from $Y'$ to $Y$.  

Proof:  For simplicity, assume it's a knot $K$ and the surgery coefficient is an integer $n$.  (It's not hard to push this argument to the general case, just more annoying to write down.)  Now, attach a 2-handle to $Y$ along $K$ with framing $n$.  Call the 2-handle cobordism $W$.  Since $K$ is nullhomotopic, $W$ is homotopy equivalent to $Y \vee S^2$ and so we can retract onto $Y$.  So, here's the map from $Y'$ into $Y$: include $Y'$ into $W$, do the homotopy / retraction onto $Y$.  Since $[Y'] = [Y]$ in $W$ and the retraction preserves $H_3$, this becomes a degree one map.  

Pretty cool right!  An alternative proof is to double the 2-handle cobordism and use Kirby calculus to prove that this four-manifold is $Y \times I \# S^2 \times S^2$ or $Y \times I \# \mathbb{C}P^2 \# \overline{\mathbb{C}P^2}$, depending on parity of the surgery coefficient.  Then $Y'$ includes into this manifold, which we can collapse to $Y \times I$ and then project down to $Y$.  This is again a degree one map.  (Now you can kind of see how Gadgil's work comes from.)

As you might expect, a lot of questions about maps between 3-manifolds can be related to the special structures of 3-manifold groups (e.g. being Hopfian).  I'll mention briefly that there's also a whole story about maps between knot groups as well which is interesting in its own right as well.  For example, there's an old conjecture of Simon (Problem 1.12(B) in the Kirby list) that if there is a $\pi_1$-surjection from $S^3 - K'$ to $S^3- K$, then the genus of $K'$ at least that of $K$?  (It's not too hard to prove that there is a similar inequality for the degrees of the Alexander polynomials for example.)  I'll end by saying that in parallel with what's mentioned above, Agol and Liu proved that a given knot group surjects at most finitely many knot groups (this was Problem 1.12(C) in the Kirby list).