Non-surjective map implies zero degree for manifolds with boundary 
Let $M$ and $N$ be compact, connected, orientable, topological $n$-manifolds with non-empty boundary, and let $$f : (M, \partial M) \to (N , \partial N) $$ be a continuous map of pairs.
Then, we can define the degree of $f$ as the integer $d \in \mathbb{Z}$ for which $$f_*([M]) = d[N], $$ where $[M] \in H_n(M, \partial M; \mathbb{Z})$ and $[N] \in H_n(N, \partial N; \mathbb{Z})$ are relative fundamental classes.



I am trying to prove/disprove the statement that if $f$ is not surjective, then the degree of $f$ must be zero. I think this is true, however, I am not sure how to prove this.


If $M$ and $N$ would not have boundary, the usual way to prove such a statement would be to use the characterization of the degree by local degrees and regular values. Is there a similar result in the case of non-empty boundary?
I also though about the idea of the proof in the case of spheres: if $y \in N$ is a point not in the image of $f$, then $f$ factors through as a map $$M \to N \setminus \{y\} \hookrightarrow N, $$ but I do not know why this would be the zero map.
 A: Let $f: (M,\partial M) \rightarrow (N,\partial N)$ be continuous nonsurjective. Its image is still compact, so that there is a $y \in N \backslash (f(M) \cup \partial N)$. Write $N’=N \backslash \{y\}$. Let $y \in B \subset N \backslash \partial N$ be a small ball around $y$.
If $H_n(N’, \partial N; \mathbb{Z})=0$, then the conclusion follows.
Now, we have a relative homology LES (with coefficients in $\mathbb{Z}$): $$H_{n+1}(N,N’) \rightarrow H_n(N’,\partial N) \overset{f}{\rightarrow} H_n(N,\partial N) \rightarrow H_n(N,N’).$$
But by excision and the relative homology LES (as $B$ is contractible), $H_k(N,N’) \cong H_k(B,B \backslash \{y\}) \cong \tilde{H}_{k-1}(B \backslash \{y\})$ ($\tilde{H}$ denotes reduced homology). In particular, this group is zero for $k=n+1$ and isomorphic to $\mathbb{Z}$ for $k=n$, so $H_n(N’,\partial N)=\ker{f}$.  But $H_n(N,\partial N)$ is isomorphic to $\mathbb{Z}$ too.
The goal is thus to show that $f$ is injective, ie nonzero. But this follows from the definition of the relative fundamental class (see the discussion in Hatcher’s Algebraic Topology [online], just above III.3.43).
A: If $f$ is not surjective, then the degree is necessarily zero.  The way I'll prove this is a follows.  First, I'll show how to turn $f$ into a new map $f'$ between closed oriented manifolds (i.e., they have no boundary) in such a way that $\deg f = \deg f'$.  The process will obviously take a non-surjective $f$ to a non-surjective $f'$.  But for closed manifolds, we already know that $\deg f' = 0$ if $f'$ is not surjective.
So, let's construct $f'$ and the closed manifolds.  I will assume $M$ is oriented, and I'll let $-M$ denote $M$ with the opposite orientation.
Form the space $M' := M\coprod_{\partial M} -M$, where we glue $M$ to $-M$ along their common boundary $\partial M$ via the identity map.  Note that $M'$ is a topological manifold without boundary, and is naturally oriented.  Similarly, form the space $N'$.  Note that $M'$ contains a canonical copy of $M$ and a canonical copy of $-M$, and the the two boundaries, $\partial M$ and $\partial(-M)$ are identified, forming a codimension $1$ embedded submanifold of $M'$.  Similar statements, of course, also apply to $N'$.
Then $f$ induces a map $f':M'\rightarrow N'$ defined as follows.  For $x\in M$, define $f'(x) = f(x) \in N\subseteq N'$.  For $x\in -M$, also define $f'(x) = f(x)\in -N\subseteq N'$.  For $x\in \partial M = \partial (-M)$ (in $M')$, these two definitions agree, so from the pasting lemma, $f'$ is continuous.  Note that if $f$ happens to not be surjective, say, missing a point $n\in N$, then $f'$ also misses $n\in N\subseteq N'$, so $f'$ is not surjective either.
Fixing $x\in \partial M\subseteq M'$ and noting that $f(x)\in \partial N\subseteq N'$ we may view $f'$ as either a map $(M',\partial M)\rightarrow (N',\partial N)$ or as a map $(M', x)\rightarrow (N', f(x))$.
We obtain a commutative diagram:
\begin{align*} (M', x) & \xrightarrow{f'} & (N',f(x))\\
\downarrow & & \downarrow\\
(M', \partial M) & \xrightarrow{f'} & (N', \partial N)\\
\uparrow & & \uparrow\\
(M,\partial M) & \xrightarrow{f} & (N,\partial N)\\
\end{align*}
Let's compute $H_n$ of this diagram, where $n = \dim M$.  To that end, note that $H_n(M',\partial M)\cong \mathbb{Z}^2$.  Indeed, the quotient map $(M',\partial M)\rightarrow (M'/\partial M, \partial M/\partial M)$ induces an isomorphism since $(M',\partial M)$ is a good pair.  But $M'/\partial M$ is homeomorphic to the one point union $(M/\partial M)\vee (-M/\partial M)$.   Thus $$H_n(M',\partial M) \cong H_n(M/\partial M)\oplus H_n(M/\partial M)\cong H_n(M,\partial M)\oplus H_n(M,\partial M)\cong \mathbb{Z}\oplus \mathbb{Z}.$$
So, taking $H_n$ of the diagram, we obtain the following diagram: \begin{align*} \mathbb{Z} & \xrightarrow{\deg f'} & \mathbb{Z}\\
\downarrow & & \downarrow\\
\mathbb{Z}^2 & \xrightarrow{f'_\ast} & \mathbb{Z}^2\\
\uparrow & & \uparrow\\
\mathbb{Z} & \xrightarrow{\deg f} & \mathbb{Z}
\end{align*}
Proposition:  We have $\deg f = \deg f'$.
Proof:
From the construction, the upward maps send $1$ to $(1,0)$.  Starting in the bottom left corner and moving right then up, we find $1\mapsto \deg f \mapsto \deg f\cdot (1,0)$.
Again starting in the bottom left, but this time moving up then right, we get $1\mapsto (1,0)\mapsto f_\ast(1,0)$, so we see that $f_\ast(1,0) = \deg f\cdot (1,0)$.
If we repeat the construction of the diagram with the bottom spaces replaced by $(-M,\partial M)$ and $(-N,\partial N)$, the upward map sends $1$ to $(0,1)$.  Now the same proof shows that $f_\ast(0,1) = \deg f\cdot (0,1)$.
It now follows that $f_\ast$ is simply multiplication by $\deg f$.
Now let's look at the downward map.  This will send $1$ to $(1,1)$.  Starting with a $1$ in the top left and moving down then right, we find $1\mapsto (1,1)\mapsto \deg f\cdot (1,1)$.  On the other hand, starting with a $1$ in the top left, and then moving right and then down, we find $1\mapsto \deg f' \mapsto \deg f'\cdot (1,1)$.
So, we see $\deg f\cdot (1,1) = \deg f'\cdot (1,1)$, so $\deg f = \deg f'$ as claimed.  $\square$.
