TITLE: Existence of non-null-homotopic map from $M^n$ to $S^{n-1}$
QUESTION [14 upvotes]: Let $M^n$ be compact, connected, oriented $n$-dimensional smooth manifold without boundary, the Hopf degree theorem states that the homotopy class of continuous maps from $M^n$ to $S^n$ is classified by its degree. What about continuous map from $M^n$ to $S^{n-1}$ (assume $n>0$) ? I am mostly interested in the case $n>2$.
More specially, when does there exist a non null-homotopic map in the case $n=3$? If $H^2(M,\Bbb Z) \not=0$ then it's true as $[M,\Bbb{C}P^{\infty}]=H^2(M,\Bbb Z)$ and $\Bbb CP^1 \cong S^2$. This is not necessary as $M=S^3$ shows.

REPLY [12 votes]: For $n=3$, there are good general results about $[X,S^2]$ and $[X,S^3]$ (for arbitrary spaces $X$) in the paper The principal fibration sequence and the second cohomotopy set by Laurence Taylor.
For $n\geq 4$ I claim that there is an exact sequence
$$ H^{n-2}(M;\mathbb{Z}) \xrightarrow{\text{Sq}^2\circ\rho} H^n(M;\mathbb{Z}/2) \simeq\mathbb{Z}/2 \to [M,S^{n-1}] \to H^{n-1}(M;\mathbb{Z}) \to 0 $$
The first map here is the composite of the reduction map $\rho\colon H^{n-2}(M;\mathbb{Z})\to H^{n-2}(M;\mathbb{Z}/2)$ with the Steenrod operation $\text{Sq}^2\colon H^{n-2}(M;\mathbb{Z}/2)\to H^n(M;\mathbb{Z}/2)$.  It could be analysed further using the universal coefficient theorem and the theory of Wu classes, but I will not go into that here.
To prove the claim, standard calculations show that for $n\geq 4$ we have $\pi_{n-1}(S^{n-1})=\mathbb{Z}\iota$ and $\pi_n(S^{n-1})=\mathbb{Z/2}\eta$ and $\pi_{n+1}(S^{n-1})=\mathbb{Z}/2\eta^2$.  Also, the homotopy groups of the connective real $K$-theory spectrum $kO$ are $(\mathbb{Z}\iota,\mathbb{Z}/2\eta,\mathbb{Z}/2\eta^2,0,\dotsc)$.  Now let $B$ be the $(n-1)$'th space in the $\Omega$-spectrum for $kO$, so $[X,B]=kO^{n-1}(X)$ for all spaces $X$.  Let $F$ be the fibre of the unit map $S^{n-1}\to B$.  From the long exact sequence of the fibration we find that $\pi_i(F)=0$ for $i\leq n+1$.  By induction over the cells of $X$, we deduce that the map $[X,S^{n-1}]\to [X,B]$ is bijective whenever $X$ is a CW complex of dimension at most $n$.  In particular, we can take $X=M$ to see that $[M,S^{n-1}]=kO^{n-1}(M)$, so we now have a calculation in stable homotopy theory. 
I will formulate the rest of the argument in terms of the Atiyah-Hirzebruch spectral sequence, but it could be reformulated in terms of obstruction theory if desired.
The spectral sequence has the form
$$ E_2^{pq} = H^p(M;kO^q) \Longrightarrow kO^{p+q}(M), $$
with differentials $d_r\colon E_r^{pq}\to E^{p+r,q-r+1}$.  Here $kO^q$ is the same as $\pi_{-q}kO$, so it is trivial unless $q\leq 0$, so the spectral sequence is concentrated in the fourth quadrant.  Also, the cohomology of $M$ is concentrated in degrees zero to $n$, so the spectral sequence is concentrated in columns zero to $n$.  The target group $kO^{n-1}(M)$ has a filtration with quotients $E_\infty^{p,n-1-p}$.  The only possible nonzero terms are $E_\infty^{n-1,0}$ (which is a subquotient of $E_2^{n-1,0}=H^{n-1}(M;\mathbb{Z})$) and $E_\infty^{n,-1}$ (which is a subquotient of $E_2^{n,-1}=H^n(M;\mathbb{Z}/2\eta)$).  We therefore have a short exact sequence 
$$ E_\infty^{n,-1} \to kO^{n-1}(M) \to E_\infty^{n-1,0} $$
Now recall that $d_r$ has bidegree $(r,1-r)$.  It follows that all differentials starting at $(n-1,0)$ or $(n,-1)$ have endpoint in the region $p>n$ where the spectral sequence is zero.  Similarly, all differentials ending at $(n-1,0)$ or $(n,-1)$ have starting point in the region $q>0$ where the spectral sequence is zero, with the exception of the differential
$$ d_2\colon H^{n-2}(M;\mathbb{Z}) = E_2^{n-2,0} 
    \to E_2^{n,-1} = H^n(M;\mathbb{Z}/2\eta).
$$
Thus, $E_\infty^{n,-1}$ is the cokernel of the above differential, whereas $E_\infty^{n-1,0}$ is just the same as $E_2^{n-1,0}=H^{n-1}(M;\mathbb{Z})$.  By considering the Posnikov tower of $kO$, one can also check that the nontrivial differential is the composite of the reduction map $\rho\colon H^{n-2}(M;\mathbb{Z})\to H^{n-2}(M;\mathbb{Z}/2)$ with the Steenrod operation $\text{Sq}^2\colon H^{n-2}(M;\mathbb{Z}/2)\to H^n(M;\mathbb{Z}/2)$.