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Suppose $M$ is a connected compact topological manifold.

If you consider a closed connected 1-dimensional manifold $X$, then it's homeomorphic to a circle $X\cong \mathbb{S}^1$. Let's denote $[X,M]$ the space of classes of continous functions $f:X\rightarrow M$ up to homotopy. It's easy to see that: $$ [X,M]\cong\pi_1(M) $$

My question is: if you have a closed connected and orientable 2-dimensional manifold $X$ (which can be classified by its genus $g$). Can $[X,M]$ be fully characterized by $g$, $\pi_1(M)$ and $\pi_2(M)$? If not, are there two manifolds $M_1,M_2$ with identical fundamental and second-homotopy groups but such that $[X,M_1]\neq [X,M_2]$ for some genus $g$?

In some sense, if we define $\pi_2^g(M)=[X,M]$, then it generalizes $\pi_2(M)$ with $\pi_2^0=\pi_2$. My question can also be thought as asking: do we gain any new topological information about $M$ by considering $\pi_2^g$?

Diana Pestana
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  • A related question is this one, but I think my question is easier in some sense. I am also accepting partial answers for some $g\neq 0$. – Diana Pestana Apr 18 '25 at 05:01
  • Yeah, thanks for warning me. I just fixed those. – Diana Pestana Apr 18 '25 at 06:07
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    Already the claim that you say "is easy to see" is false. – Moishe Kohan Apr 18 '25 at 08:14
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    This answer computes the set of based homotopy classes $[T^2, M]$. I think this should give counterexamples by finding manifolds with a suitable action of $\pi_1$ on $\pi_2$. I don't know an example off the top of my head, though. – Derived Cats Apr 19 '25 at 16:03
  • Oh, so if one can generalize that result for all genus. Then $[X,M]$ deppends on $\pi_1(M), \pi_2(M)$ but also on the $\pi_1(M)$-action on $pi_2(M)$? It's a bit different from what I expected, but that basically answers my original motivation for that question! – Diana Pestana Apr 19 '25 at 18:28
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    As Moishe Kohan's comment alludes to, you need to take basepoints into account to identify $[X, M]$ with $\pi_1(M)$. Even if you do that when defining $\pi_2^g$, you won't get a group for $g > 0$. – Michael Albanese Apr 21 '25 at 10:53

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