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Let $D(M,N)$ be the set of all possible degrees of maps from $M$ to $N$, $M_1\#M_2$ the connected sum of $M_1$ and $M_2$.

  1. Can $D(M_1\#M_2,N)$ be calculated in terms of $D(M_1,N)$ and $D(M_2,N)$?
  2. Can $D(N,M_1\#M_2)$ be calculated in terms of $D(N,M_1)$ and $D(N,M_2)$?

Here all manifolds are assumed to be of the same dimension $d \ge 3$. I'm especially interested in the case $d=3$.

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  • $\begingroup$ Have you given the question much thought? I think (2) can be readily answered by considering a few examples. $\endgroup$ – Ryan Budney Apr 13 '17 at 20:28
  • $\begingroup$ @RyanBudney I immediately see a subset relation for 1, but not the full answer. Question 2 looks rather tricky, unless you are implying the answer is "no". $\endgroup$ – Sebastian Goette Apr 14 '17 at 12:33
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I think that looking at surfaces one can see that there is no nice relation between both sides. Let $\Sigma_g = \#^g S^1 \times S^1$ be the oriented closed surface of genus $g$. Then one can show, using simplical volumes aka the Gromov norm (see http://www.map.mpim-bonn.mpg.de/Simplicial_volume), which satisfies $||\Sigma_g|| = 4g-4$, that for $g,h >1$, we have $$D(\Sigma_g,\Sigma_h) = \left\{0,\pm 1, \dots, \pm \left\lfloor \frac{g-1}{h-1}\right\rfloor\right\}.$$ In particular we have $D(\Sigma_g,\Sigma_h) = \{0\}$ for $g < h$. I leave it to you to conclude from this that there is no nice relation (just choose $g$ and $h$ in the right way...).

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Here's an elementary observation that gives a relation between the three sets in both cases.

Note that $M_1\# M_2$ admits degree one maps $\alpha : M_1\# M_2 \to M_1$ and $\beta : M_1\#M_2 \to M_2$ (just crush the other summand).

If $f : M_1 \to N$ has degree $d$, then $f\circ\alpha : M_1\# M_2 \to N$ also has degree $d$. Therefore $D(M_1, N) \subseteq D(M_1\# M_2, N)$ and likewise $D(M_2, N) \subseteq D(M_1\# M_2, N)$ so

$$D(M_1, N)\cup D(M_2, N) \subseteq D(M_1\#M_2, N).$$

If $g : N \to M_1\# M_2$ has degree $d$, then $\alpha\circ g : N \to M_1$ also has degree $d$. Therefore $D(N, M_1\# M_2) \subseteq D(N, M_1)$ and likewise $D(N, M_1\# M_2) \subseteq D(N, M_2)$ so

$$D(N, M_1\# M_2) \subseteq D(N, M_1)\cap D(N, M_2).$$


In general, the inclusions are strict as can be seen by using Jens Reinhold's answer: take $M_1 = M_2 = N = \Sigma_2$ for example.

There are also counterexamples in higher dimensions.

Let $d \geq 3$. If $M$ and $N$ are closed, connected, oriented manifolds of dimension $d$, then $\|M\# N\| = \|M\| + \|N\|$; note, this is not the case in dimension two (the torus has Gromov norm zero, but higher genus surfaces do not).

Suppose $M$ has dimension $d$ and $\|M\| \neq 0$. If $f : M \to M\# M$, then

$$\|M\| \geq |\deg f|\|M\# M\| = 2|\deg f|\|M\|$$

so $f$ has degree zero, i.e. $D(M, M\#M) = \{0\}$. Note however that $1 \in D(M\# M, M\#M)$ so

$$D(M, M\#M)\cup D(M, M\# M) \subsetneq D(M\#M, M\#M).$$

Also $1 \in D(M, M)$ so

$$D(M, M\#M) \subsetneq D(M, M)\cap D(M, M).$$

Gromov proved that the Gromov norm of a closed, oriented, hyperbolic $n$-manifold is a non-zero multiple (depending only on $n$) of its volume. In particular, such manifolds have non-zero Gromov norm so they can be used to construct counterexamples in any dimension greater than or equal to $2$.

These counterexamples seem to suggest that the answer to both questions of your questions is no.

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One has $D(M_1\#M_2,N) \supseteq D(M_1,N)+D(M_2,N)$ (here, of course, I'm assuming $N,M_1,M_2$ are all connected and oriented as well as having the same dimension). One way of saying it is that there is an oriented $d+1$-cobordism $W$ from $M_1 \sqcup M_2$ to $M_1\#M_2$ by attaching a 1-handle to $(M_1\sqcup M_2)\times I$. Given maps $f_i: M_i\to N$ of degree $d_i$, there is no obstruction to extend $f_1\sqcup f_2$ to $F:W\to N$ since $N$ is connected. Then the degree of $F_{|M_1\#M_2}$ will be $d_1+d_2$.

In case $\pi_{d-1}(N)=0$, one will have $D(M_1\#M_2,N) = D(M_1,N)+D(M_2,N)$. The point is that a map $f:M_1\#M_2\to N$ will extend to a map $F:W\to N$, since one may attach the $d$-handle along the connect sum sphere $S^{d-1}\subset M_1\#M_2$.

For $d=3$, one has $\pi_2(N)\neq 0$ iff $N=S^1\times S^2$ or $N$ is a non-trivial connect sum $N=N_1\#\cdots \# N_k$, where $N_i$ is prime. In the first case, equality will hold again, because the separating connect-sum sphere $S^2\subset M_1\#M_2$ cannot map non-trivially to an essential 2-sphere in $S^2\times S^1$ since it is homologically trivial. So one may extend over $W$ again.

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