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Consider the set of continuous maps $f$ taking the 2-torus $T^2$ to a manifold $M$, up to homotopy equivalence. Picking a pair of generators of $\pi_1(T^2)$, call them $a$ and $b$, we can map them to elements of $\pi_1(M)$, so the equivalence classes of maps $f$ are at the very least distinguished by elements of $\pi_1(M)^2$.
But additionally, given a point $x$ in $T^2$ and fixing its image $y$ in $M$, for any element $g$ of $\pi_2(M)$ we can pick a representative based at $y$, then find two maps $f$ and $f'$ differing only in a small neighborhood of $x$ such that their images in that neighborhood differ by that representative of $g$. (At least, this is my understanding -- it's just a matter of cut-and-pasting a sphere at the point $y$.) This gives us a set of classes of maps $f$ which are distinguished by elements of $\pi_1(M)^2 \times \pi_2(M)$.

Is this sufficient to characterize the equivalence classes of maps up to homotopy? Or is there something else I'm missing?

More generally, given two (finite-dimensional) manifolds $M_1$ and $M_2$, is all the data necessary to describe the homotopy-equivalence classes of maps $f: M_1 \rightarrow M_2$ encoded in the homotopy groups of $M_1$ and $M_2$, or is more data needed?

Note: I am assuming connected manifolds here.

Ben Steffan
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Craig
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    Your argument is a bit too vague for me to be convinced. How do you know the element $g$ is encoded by some $f,f'$? What does it mean on a manifold, without an Abelian structure, that $f-f'=g$? This "differ"? I guess you want to work in local coordinates, but how do you know $g$ is homotopic to something landing in a local coordinate chart? In fact, it never will be - so long as $g$ is nontrivial. – FShrike Aug 28 '24 at 11:20
  • Look, consider $g$ in $\pi_2(M)$. You can imagine this as a non-contractible sphere embedded in $M$; you can always define an arbitrary base point $y \in M$ to which $g$ is attached (more precisely, there is a submanifold of $M$ containing $y$ in the homotopy class of $g$). Now cut $g$ along an arbitrary closed curve with trivial homotopy in $M$ (e.g. take a small neighborhood of $y$) and define the two parts of $g$ obtained by $g_1$ (contractible in $M$) and $g_2$. Let $f$ map a neighborhood of $x$ to $g_1$ and $f'$ map the same neighborhood to $g_2$. – Craig Aug 28 '24 at 13:15
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    You might want to think about the following setup instead: $S^1 \to S^1 \vee S^1 \to S^1 \times S^1 \to S^2$ is a (based) Puppe sequence, so you obtain an exact sequence $[S^2, M]* \to [S^1 \times S^1, M]* \to [S^1 \vee S^1, M]* \to [S^1, M]*$ of pointed sets. 3 out of 4 terms here are just homotopy groups (and you can continue the sequence further to the left if need be), so you might have a shot. – Ben Steffan Aug 28 '24 at 14:00
  • What? No. $\pi_2(M)$ is literally defined as the homotopy equivalence classes of maps $g: [0,1]^2 \rightarrow M$ taking the boundary of a square to a fixed point (say $y$) in $M$ (look in Wikipedia for example). For any given $f$ and $x$ as above, we can find $\hat{f}$ homotopically equivalent to $f$ with a small square containing $x$ mapping to $y$. We can then define $f'$ such that it is identical to $\hat{f}$ outside the square, and equivalent to $g$ inside the square. Are you objecting to calling $g$ a non-contractible sphere? That's reasonable; really thats only true if $g$ is a generator – Craig Aug 28 '24 at 14:17
  • Ben Stefan certainly knows the definition of $\pi_2$. This question would be much improved if you carefully stated some points you’re making and proved (or offered more detailed proof-sketches for) them. – FShrike Aug 28 '24 at 22:03
  • My comment was in response to a version of his comment that has since disappeared. I believe that my new comment shows how $\pi_2$ comes into play. – Craig Aug 29 '24 at 03:04
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    Yes, this is solved here https://math.stackexchange.com/questions/36488/how-to-compute-homotopy-classes-of-maps-on-the-2-torus – Nick L Aug 29 '24 at 13:49
  • Thank you Nick L, that is exactly what I was looking for. Do you know the answer to the more general question? Specifically, given $M_1$ and $M_2$ finite-dimensional connected manifolds, is $[M_1, M_2]_*$ determined by the homotopy groups of $M_1$ and $M_2$? – Craig Aug 29 '24 at 16:26
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    @Craig The answer to your latter question is negative. E.g. $S^2\times\mathbb{R}P^3$ and $\mathbb{R}P^2\times S^3$ are closed $5$-manifolds with the same homotopy groups, but they have different $5$-th cohomology since one is orientable and the other is not, hence different sets of homotopy classes of maps into $S^5$ by Hopf's theorem. – Thorgott Aug 30 '24 at 02:00

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