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How to compute $[T^{2},T^{2}]$ the set of homotopy class of continuous maps $f:T^{2}\longrightarrow T^{2}$?

Thanks.

R.Bradley
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1 Answers1

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Since $T^2=S^1\times S^1$, you get $$ [T^2,T^2]=[T^2,S^1]\times [T^2,S^1]. $$ So, the problem reduces to finding $[T^2,S^1]$. Since $S^1$ is the first Eilenberg-Maclane space, we have that $$ [T^2,S^1]\cong H^1(T^2;\mathbb{Z})\cong \mathbb{Z}\oplus\mathbb{Z}. $$

J126
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    That's a great answer, a small modification would be to notice that $T^{2}$ itself is an Eilenberg-Maclane space! – Piotr Pstrągowski Oct 31 '13 at 17:06
  • How you compute the map $deg:[T^{2},T^{2}]\longrightarrow\mathbb{Z}$? – R.Bradley Oct 31 '13 at 17:21
  • @R.Bradley What is your definition for $\deg$ when the space is not a sphere? – J126 Oct 31 '13 at 17:55
  • Let be $F:X\longrightarrow Y$ ($X,Y$ compact, connected manifolds) then $F^{*}:H^{n}(X)\longrightarrow H^{n}(Y):[X]\mapsto d\cdot[Y]$ we say that $d=deg(F)$. – R.Bradley Oct 31 '13 at 18:05
  • Since $H^2(T^2)\cong\mathbb{Z}$, given a map $f\in [T^2,T^2]$, you would need to compute $f^*:\mathbb{Z}\rightarrow \mathbb{Z}$. Using cellular cohomology would probably be the most convenient method. But, you need to have a specific $f$ in mind. – J126 Oct 31 '13 at 18:09
  • And other method instead of cellular cohomology? – R.Bradley Oct 31 '13 at 18:17
  • @R.Bradley There are other methods. Again, the best method would depend on the $f$ of which you are trying to compute the degree. – J126 Oct 31 '13 at 19:03
  • Other question: If possible compute $[T^{2},T^{2}]$ using De Rham cohomology? – R.Bradley Oct 31 '13 at 19:20
  • @R.Bradley That is something you should ask as another question. We shouldn't converse in the comments for so long. – J126 Oct 31 '13 at 20:12
  • This answer shows that $[T^2,T^2] \cong \mathbb{Z}^4$, and it is not hard to show that a representative of the class $(a,b,c,d) \in \mathbb{Z}^4$ is given by the function $T^2 \to T^2$, $(w,z) \to (w^az^b, w^cz^d)$ (where I think of $T^2$ as $S^1 \times S^1$, and of $S^1$ as the unit norm complex numbers). That map has degree $ab+cd$, by counting pre-images. Therefore the map $deg : \mathbb{Z}^4 \cong [T^2, T^2] \to \mathbb{Z}$ is given by $(a,b,c,d) \mapsto ab+cd$, @R.Bradley. – Omar Antolín-Camarena Nov 01 '13 at 17:42