Homotopic maps to $S^n$ I'm working through a proof that, given an oriented compact (connected) $n$-manifold $M$ with boundary, any two continuous maps $f,g:M\to S^n$ are homotopic. The proof uses the double of $M$, which is obtained by gluing together two copies of $M$ at the boundary to obtain a smooth, compact, closed, oriented $n$-manifold; call it $M'$. Let $\pi:M'\to M$ be the natural projection and $\iota: M\to M'$ the natural embedding. Then, $deg (f\circ \pi)=0$ and so $f=f\circ \pi\circ \iota$ is homotopic to a constant map. 
Why is $deg(f\circ \pi)=0$? 
 A: Here's a proof in terms of homology.  Let $x$ be an $n$-chain on $M$ that is a fundamental class of $(M,\partial M)$ (i.e., $x$ is a cycle relative to $\partial M$ and $[x]\in H_n(M,\partial M)$ is the generator corresponding to the orientation on $M$).  Let $i,j:M\to M'$ be the two inclusion maps.  Then I claim that $y=i_*(x)-j_*(x)$ is a fundamental class for $M'$.  Indeed, since $\partial x$ is contained in $\partial M$, $\partial i_*(x)=\partial j_*(x)$, so $y$ is a cycle.  Now note that $H_n(M',j(M))\cong H_n(M,\partial M)$ by excision and that the image of $[y]$ in $H_n(M',j(M))\cong H_n(M,\partial M)$ is just $[x]$, which is a generator of $H_n(M,\partial M)$.  If $\varphi:\mathbb{Z}\to\mathbb{Z}$ is a homomorphism and $\varphi(n)=1$, we must have $n=\pm 1$.  Applying this to the map $\mathbb{Z}\cong H_n(M')\to H_n(M,\partial M)\cong\mathbb{Z}$, we conclude that $[y]$ must be a generator of $H_n(M')$.
So to compute the degree of $f\circ \pi$, we can just compute $(f\circ\pi)_*([y])\in H_n(S^n)$.  But $\pi_*i_*=\pi_*j_*$ so $f_*\pi_*(y)=f_*\pi_*i_*(x)-f_*\pi_*j_*(x)=0$ (as a chain, not just as a homology class!).  Thus $\deg(f\circ \pi)=0$.
