Let $M$ be a connected, compact, oriented manifold of dimension $n<7$. If any two maps $M \to M$ having equal degrees are homotopic, must $M$ be diffeomorphic to the $n$sphere?

5$\begingroup$ Of course not. Consider a homotopysphere for example. $\endgroup$ – Ryan Budney Jul 21 '13 at 7:01

2$\begingroup$ There's infinitely many examples that aren't even homotopy spheres. Just take any manifold so that the group of homotopyclasses of homotopy selfequivalences is trivial. For example, a hyperbolic manifold with trivial isometry group. $\endgroup$ – Ryan Budney Jul 21 '13 at 7:25

2$\begingroup$ @Ryan: As I am sure you know, in dimensions less than seven and different from four, any homotopy sphere is diffeomorphic to the standard sphere; in dimension four, the smooth Poincaré conjecture is unresolved. $\endgroup$ – Ricardo Andrade Jul 21 '13 at 7:47

1$\begingroup$ It follows from Mostow rigidity. There's plenty of symmetryless hyperbolic 3manifolds. Every finite group is the isometry group of a hyperbolic 3manifold, and the trivial group is the isometry group of infinitely many. $\endgroup$ – Ryan Budney Jul 21 '13 at 10:22

3$\begingroup$ @Ryan: in the hyperbolic manifold example, how about nonhomotopic zerodegree maps? $\endgroup$ – Sergei Ivanov Jul 21 '13 at 18:21
$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\CC}{\mathbb{C}}$A simple counterexample is given by $M = \CC P^3$.
Recall first that the cohomology ring of $\CC P^3$ is a truncated polynomial algebra: $$ H^\ast(\CC P^3;\ZZ) = \ZZ[x]/(x^4) $$ with $x \in H^2(\CC P^3;\ZZ)$. The set of homotopy classes of selfmaps of $\CC P^3$ is $$ [\CC P^3,\CC P^3] = [\CC P^3,\CC P^\infty] = H^2(\CC P^3;\ZZ) = \ZZ $$ where a selfmap $f$ of $\CC P^3$ is taken to the unique integer $k\in\ZZ$ such that $f^\ast x = kx$. The first isomorphism above, $[\CC P^3,\CC P^3] = [\CC P^3,\CC P^\infty]$, follows by cellular approximation from:
 the source $\CC P^3$ is a CWcomplex of dimension six;
 the target $\CC P^3$ is the skeleton of dimension seven of the usual CWstructure on $\CC P^\infty$.
Consequently, a selfmap $f:\CC P^3 \to \CC P^3$ corresponding to the integer $k \in \ZZ$ has degree $k^3$: we have $f^\ast x = kx$, therefore $$ f^\ast(x^3) = (f^\ast x)^3 = (kx)^3 = k^3 x^3 $$ Since $k\mapsto k^3$ is an injective map $\ZZ\to\ZZ$, any two selfmaps of $\CC P^3$ which have the same degree are homotopic. On the other hand, $\CC P^3$ is not diffeomorphic, or even homotopy equivalent, to $S^6$.
Further example added later:
$\newcommand{\To}{\longrightarrow}$$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\QQ}{\mathbb{Q}}$$\newcommand{\RR}{\mathbb{R}}$Another counterexample is given by any real projective space $\RR P^n$ of odd dimension $n>1$. So we get the counterexamples $\RR P^3$ and $\RR P^5$ in dimensions less than seven.
I have failed to find any reference with a proof that two selfmaps of $\RR P^n$ which have the same degree are homotopic, although this result is stated without proof as theorem 2.1 in this article (McGibbon, Selfmaps of projective spaces, Transactions of the A.M.S. 271 (1), 1982, pages 325346). For completeness, I will give a proof below. The proof is more homotopy theoretical than for the case of $\CC P^3$ above, yet also uses fairly standard techniques.
[The rest of this answer is rather long, and you may ignore it if you are not interested in the proof for $\RR P^n$.]
Proof that two selfmaps of $\RR P^n$ (with $n>1$ odd) are homotopic if they have the same degree
Assume that $n$ is an odd integer with $n>1$. I will actually prove the stronger result that two basepointpreserving selfmaps of $\RR P^n$ which have the same degree are homotopic via a basepointpreserving homotopy. More precisely, I will prove that the map which takes (the pointed homotopy class of) a selfmap to its degree $$ \deg : [\RR P^n,\RR P^n]_\ast \To \ZZ $$ is a bijection. Let us start with a lemma which will be useful later on.
Lemma: A map $f:\RR P^n \to \RR P^n$ (with $n>1$) has even degree if and only if $f$ induces the trivial map on fundamental groups.
Proof: This proof involves some cohomological computations similar to the ones done earlier for $\CC P^3$. Note that $H^\ast(\RR P^n;\ZZ/2) = \ZZ/2[x]/(x^{n+1})$ where $x$ is in degree $1$. Since $\pi_1(\RR P^n)=\ZZ/2$, a selfmap $f$ of $\RR P^n$ induces the trivial homomorphism on $\pi_1$ if and only if it induces the trivial homomorphism on $H^1(;\ZZ/2)$, i.e. if and only if $f^\ast x = 0$. Letting $f^\ast x = k x$ with $k\in\ZZ/2$, then $f^\ast(x^n) = k^n x^n = k x^n$, so $k = \deg f \mod 2$. The result follows. ■
A cofibre sequence involving $\RR P^n$. Our starting point is the following description of $\RR P^n$. The space $\RR P^n$ is obtained by attaching a $n$cell to $\RR P^{n1}$ whose attaching map is the standard covering map $\pi : S^{n1} \to \RR P^{n1}$. In short, $\RR P^n$ is the homotopy cofibre of the map $\pi$. The associated homotopy cofibre sequence — extended to the right by one term — is: $$ S^{n1} \overset{\pi}{\To} \RR P^{n1} \overset{i}{\To} \RR P^n \overset{q}{\To} S^n $$ Here, the first map $\pi$ is the standard quotient map, the second map $i$ is the inclusion, and the third map $q$ collapses $\RR P^{n1}$ to a point.
The associated exact sequence in homotopy. Applying the functor $[,\RR P^n]_\ast$ to the above cofibre sequence produces an exact sequence: $$ [S^n,\RR P^n]_\ast \overset{q^\ast}{\To} [\RR P^n,\RR P^n]_\ast \overset{i^\ast}{\To} [\RR P^{n1},\RR P^n]_\ast \overset{\pi^\ast}{\To} [S^{n1},\RR P^n]_\ast $$ This is an exact sequence of pointed sets. Moreover, the leftmost term is a group, and the sequence verifies a stronger equivariant exactness property at the term $[\RR P^n,\RR P^n]_\ast$ which will be explained and used later.
Calculation of the terms of the exact sequence. First, let us calculate all terms of the sequence other than $[\RR P^n,\RR P^n]_\ast$:
we have isomorphisms $[S^n,\RR P^n]_\ast = \pi_n(\RR P^n) = \pi_n(S^n) = \ZZ$ as groups (by using the covering map $S^n \to \RR P^n$ which induces an isomorphism on higher homotopy groups);
similarly, $[S^{n1},\RR P^n]_\ast = \pi_{n1}(\RR P^n) = \pi_{n1}(S^n) = 0$;
$[\RR P^{n1},\RR P^n]_\ast = [\RR P^{n1},\RR P^\infty]_\ast = \widetilde{H}^1(\RR P^{n1};\ZZ/2) = \ZZ/2$ by using cellular approximation again.
Summarizing, the previous exact sequence of pointed sets is $$ \ZZ \overset{q^\ast}{\To} [\RR P^n,\RR P^n]_\ast \overset{i^\ast}{\To} \ZZ/2 \overset{\pi^\ast}{\To} 0 $$ In particular, $i^\ast$ is a surjective function.
Description of the arrow $i^\ast$ of the exact sequence. The next step is to identify the function $i^\ast$ appearing in the exact sequence. Consider the following isomorphisms characterizing the target of $i^\ast$: $$ [\RR P^{n1},\RR P^n]_\ast = \widetilde{H}^1(\RR P^{n1};\ZZ/2) = \Hom\bigl(\pi_1(\RR P^{n1}),\ZZ/2\bigr) = \Hom\bigl(\pi_1(\RR P^{n1}),\pi_1(\RR P^n)\bigr) $$ Recall that the inclusion $i:\RR P^{n1}\to \RR P^n$ induces an isomorphism on $\pi_1$. Therefore, the function $$ i^\ast : [\RR P^n,\RR P^n]_\ast \To \Hom\bigl(\pi_1(\RR P^{n1}),\pi_1(\RR P^n)\bigr) = \ZZ/2 $$ takes $f:\RR P^n \to \RR P^n$ to $0\in\ZZ/2$ if and only if $f$ induces the trivial homomorphism on fundamental groups. The lemma at the beginning of this proof now implies that $i^\ast$ takes a map $f$ to its degree mod $2$, $\deg_2 f$.
In conclusion, the exact sequence becomes: $$ \ZZ = [S^n,\RR P^n]_\ast \overset{q^\ast}{\To} [\RR P^n,\RR P^n]_\ast \overset{\deg_2}{\To} \ZZ/2 $$ where the second arrow $\deg_2$ returns the degree mod $2$ of a selfmap of $\RR P^n$. We know from before that $\deg_2 = i^\ast$ is surjective.
Equivariant exactness property of the exact sequence. At this point, we require the special equivariant exactness property of the long exact sequence in homotopy associated to a cofibre sequence. In this case, it states:
the group $\ZZ = \pi_n(\RR P^n) = [S^n,\RR P^n]_\ast$ (with its usual group structure) acts on $[\RR P^n,\RR P^n]_\ast$;
$f$ and $g$ map to the same element via $\deg_2 : [\RR P^n,\RR P^n]_\ast \to \ZZ/2$ if and only if there exists $\alpha\in [S^n,\RR P^n]_\ast$ such that $\alpha\cdot f = g$ in $[\RR P^n,\RR P^n]_\ast$.
The action of $\alpha \in [S^n,\RR P^n]_\ast$ takes $f \in [\RR P^n,\RR P^n]_\ast$ to: $$ \alpha\cdot f : \RR P^n \overset{\sigma}{\To} S^n \vee \RR P^n \overset{\alpha+f}{\To} \RR P^n $$ where the map $\sigma$ simply "pinches a small spherical bubble off of $\RR P^n$". Note that $\deg(\alpha\cdot f) = \deg\alpha + \deg f$.
Conclusion of the proof. We are now ready to finish the proof. First we will show the injectivity of the degree function. Assume that $f$ and $g$ are pointed selfmaps of $\RR P^n$ which have the same degree. Then their degrees mod $2$ also coincide and, by the exactness property above, there exists $\alpha$ such that $\alpha\cdot f = g$. But $$ \deg f = \deg g = \deg(\alpha\cdot f) = \deg\alpha + \deg f $$ so that $\deg\alpha = 0$. It follows that $\alpha = 0$ in $\pi_n(\RR P^n)$, since elements in $\pi_n(\RR P^n) = \pi_n(S^n)$ are determined by their degree (here we require that $n$ is odd). Thus $g = \alpha\cdot f = f$ in $[\RR P^n,\RR P^n]_\ast$.
Finally, we show that any integer is the degree of some map $\RR P^n \to \RR P^n$. Recall that the function $\deg_2 : [\RR P^n,\RR P^n]_\ast \to \ZZ/2$ taking a map to its degree mod $2$ is surjective. In other words, there are selfmaps of $\RR P^n$ with even and odd degrees. Since $\deg(\alpha\cdot f) = \deg\alpha + \deg f$, the desired conclusion follows by observing that any even integer is the degree of some map $\alpha : S^n \to \RR P^n$ — which holds because the covering map $S^n \to \RR P^n$ has degree $2$ and induces an isomorphism on $\pi_n$.

1
To complete @RyanBudney's suggestion (and @Misha's comment), there are 3manifolds in the snappea census which have trivial isometry group, so in particular have no nontrivial degree 1 selfmaps, and which have trivial first betti number. If the map is degree $>0$, then this follows from Gromov's proof of Mostow rigidity (using the Gromov norm, the degree must $=\pm 1$, and the map is homotopic to an isometry). If a degree zero map from a 3manifold to itself is homotopically nontrivial, then it must surject the fundamental group of some covering space. If this covering space is finiteindex, then since the group is cofinitely Hopfian, this map must induce an isomorphism on fundamental group, and thus must be degree $\pm 1$, a contradiction. Thus, the image group must be infinite index. These groups have positive betti number, a contradiction.

$\begingroup$ Dear @Ian Agol: Very nice answer. Thank you for explaining this. If you do not mind, could you please elaborate briefly (or perhaps give a reference for) your remark between parenthesis regarding the Gromov norm? Thank you very much. Also, it may be helpful to make clear that the snappea census covers only hyperbolic manifolds (if I am not mistaken), as most people are unlikely to be familiar with snappea. $\endgroup$ – Ricardo Andrade Nov 12 '13 at 21:15

1$\begingroup$ Ricardo: Gromov norm is a homotopy invariant and is multiplicative under finite covers. $\endgroup$ – Igor Belegradek Nov 12 '13 at 23:54

1$\begingroup$ @RicardoAndrade: Here's an overview: map.mpimbonn.mpg.de/Simplicial_volume $\endgroup$ – Ian Agol Nov 13 '13 at 4:09