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Suppose I wish to find the homotopy classes of maps of $B^3 \rightarrow M$ which along the boundary are fixed by a (particular) map $f: S^2 \rightarrow M$. Take $M$ to be a closed orientable $n$-manifold; in my problem $n=9$.

What can I say about the homotopy classes of such maps?

I am sorry if this is a trivial question; at first I thought it may be $\pi_3 (M)$, but now I am not so sure...

Thanks!

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  • $\begingroup$ Do you consider homotopies that fix the restriction to $S^2$ as well? $\endgroup$
    – Daniele Zuddas
    Commented Feb 23, 2015 at 2:10
  • $\begingroup$ Yes exactly: homotopies that fix the restriction to the boundary $S^2$. $\endgroup$
    – ali elgindi
    Commented Feb 23, 2015 at 5:41
  • $\begingroup$ How can I answer this question? In particular for $M=Gr(3,6)$ real Grassmannian. $\endgroup$
    – ali elgindi
    Commented Feb 23, 2015 at 5:45
  • $\begingroup$ I appreciate any help. If I can give bounty I would! $\endgroup$
    – ali elgindi
    Commented Feb 23, 2015 at 5:46

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This set of homotopy classes is in bijective correspondence with $\pi_3(M)$. More generally, let $[B^k,X;f]$ be the set of homotopy classes of maps $B^k\to X$ that restrict to a given $f:S^{k-1}\to X$, where homotopies are also through such maps. The thing to prove is that a homotopy $F:S^{k-1}\times I\to X$ from $f$ to another map $g$ induces a bijection $[B^k,X;f]\approx[B^k,X;g]$. This is similar to the familiar basepoint-change isomorphism for homotopy groups. One defines maps $[B^k,X;f]\to[B^k,X;g]$ and $[B^k,X;g]\to[B^k,X;f]$ by putting $F$ or its inverse homotopy in a collar neighborhood of $S^{k-1}$ in $B^k$ and filling in the rest of $B^k$ with maps representing elements of $[B^k,X;f]$ or $[B^k,X;g]$, as appropriate. Then it is easy to check that these maps $[B^k,X;f]\to[B^k,X;g]$ and $[B^k,X;g]\to[B^k,X;f]$ are well-defined and are inverses of each other.

In particular, if $[B^k,X;f]$ is nonempty then $f$ extends over $B^k$ so it is homotopic to a constant map $g$, and then $[B^k,X;g]=\pi_k(X)$.

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  • $\begingroup$ Thank you very much for your help Professor Hatcher. Your books were an excellent guide during my topology classes and great reference books for the (budding) research mathematician. I understand now that homotopies of maps $B^3 \rightarrow X$ which are all fixed along the boundary $S^2$ by a specific map is equivalent to $\pi_3 (M)$. In my problem, I have two specific maps $f,g:B^3 \rightarrow X$ that agree along the boundary. All I need is that they are homotopic, not necessarily through maps all agreeing along the boundary. Would this relaxed condition make a difference? @AllenHatcher $\endgroup$
    – ali elgindi
    Commented Feb 24, 2015 at 12:13
  • $\begingroup$ If you do not require $f=g$ on $S^2$ during homotopies, then the problem becomes trivial since every map $B^3\to X$ is homotopic to a constant map. Since we may assume $X$ is path-connected, this means $f$ and $g$ are homotopic. $\endgroup$
    – Allen Hatcher
    Commented Feb 24, 2015 at 13:23
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    $\begingroup$ On the other hand, if you wish to allow only homotopies $f_t$ and $g_t$ such that $f_t=g_t$ on $S^2$ for all $t$, then such pairs $(f,g)$ are equivalent to maps $h:S^3\to X$. Homotopy classes of maps $h:S^3\to X$ without basepoint conditions are in one-to-one correspondence with orbits of the action of $\pi_1(X)$ on $\pi_3(X)$, and this set of orbits can be nontrivial. $\endgroup$
    – Allen Hatcher
    Commented Feb 24, 2015 at 13:35

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