How to compute homotopy classes of maps on the 2-torus?

Question

Let $\mathbb T^2$ be the 2-Torus and let $X$ be a topological space.

Is there any way of computing $[\mathbb T^2,X]$, the set of homotopy class of continuous maps $\mathbb T^2\to X$ if I know, for instance, the homotopy groups of $X$?

Actually, I am interested in the case $X=\mathbb{CP^\infty}$. I would like to classify $\mathbb T^1$-principal bundles over $\mathbb T^2$ (in fact $\mathbb T^2$-principal bundles, but this follows easily.)

Answer 1

As for the original question, $[\mathbb T^2,X]$ is just $\{(a,b,s)\in\pi_1(X)^2\times\pi_2(X)|ab=ba\}$ for pointed maps and $H_1(X)^2\times\pi_2(X)$ for non-pointed maps.

Indeed, parallel and meridian of $\mathbb T^2$ maps to a pair of elements of $\pi_1(X)$ and the 2-cell of the torus maps to a null-homotopy of $aba^{-1}b^{-1}$, but homotopies between trivial loop and some other null-homotopic loop can be identified with $\pi_2(X)$.

/* Moreover, this element of $\pi_2$ is well-defined: if we move $a$ by some homotopy $t$ (that can be again identified with an element of $\pi_2$), we get $s'=s+t-t$.

Cf. $[\mathbb RP^2,X]$ which gives an element of $\pi_2(X)$ well-defined only up to transforms $s'=s+2t$, i.e. only mod 2.

Actually, this kind of elementary obstruction theory (cf.) shows that for any 2-dimensional CW-complex $S$ the set of homotopy classes $[S,X]$ is in bijection with $H^1(S;\pi_1(X))\times H^2(S;\pi_2(X))$. (But in higher dimensions the situation becomes more complicated.) */

In particular, if $\pi_1(X)=0$, $[\mathbb T^2,X]=\pi_2(X)=H_2(X)$.

Answer 2

This is a good chance to advertise the paper

Ellis, G.J. Homotopy classification the J. H. C. Whitehead way. Exposition. Math. 6(2) (1988) 97-110.

Graham Ellis is referring to Whitehead's paper "Combinatorial Homotopy II", not so well read as "Combinatorial Homotopy I".

He writes:" Almost 40 years ago J.H.C. Whitehead showed in \cite{W49:CHII} that, for connected $CW$-complexes $X, Y$ with dim $X \le n$ and $\pi_i Y = 0$ for $2\le i \le \ n - 1$, the homotopy classification of maps $X \to Y$ can be reduced to a purely algebraic problem of classifying, up to an appropriate notion of homotopy, the $\pi_1$-equivariant chain homomorphisms $C_* \widetilde{X} \to C_* \widetilde{Y}$ between the cellular chain complexes of the universal covers. The classification of homotopy equivalences $Y \simeq Y$ can similarly be reduced to a purely algebraic problem. Moreover, the algebra of the cellular chains of the universal covers closely reflects the topology, and provides pleasant and interesting exercises.

"These results ought to be a standard piece of elementary algebraic topology. Yet, perhaps because of the somewhat esoteric exposition given in \cite{W49:CHII}, and perhaps because of a lack of worked examples, they have remained largely ignored. The purpose of the present paper is to rectify this situation."

Answer 3

If you want to calculate $[\mathbb T^2,\mathbb CP^\infty]$, perhaps, it's easier to use the classification of maps to $\mathbb CP^\infty$ instead: $[X,\mathbb CP^\infty]=H^2(X)$; so $[\mathbb T^2,\mathbb CP^\infty]=H^2(\mathbb T^2)=\mathbb Z$.