I have the following problem:

Given two maps $\varphi , \psi :T^2 \rightarrow T^2$, with $\varphi(p)=\psi(p)=p$ such that $\varphi_*=\psi_*$ (the induced homomorphisms on the fundamental groups based at p), then the two maps are homotopic.

Using degree theory of the circle I'm able to construct homotopies between these maps restricted to loops, but I don't see how to extend them to a continuous homotopy defined on all the torus.

Thanks in advance.

  • 1
    $\begingroup$ Hint: The torus $T^2$ is an Eilenberg-MacLane space. $\endgroup$ – Alexander Thumm May 29 '12 at 11:28
  • $\begingroup$ @alexander: thanks, but I don't know about these spaces. I know only basic results about homotopy, fundamental group and degree theory for the circle. $\endgroup$ – homeomorphism May 30 '12 at 0:04
  • $\begingroup$ Well basically a space $X$ is an Eilenberg-MacLane space of type $K(G,1)$, where $G$ is a group, if $\pi_1(X,\ast) = G$ and all other homotopy groups of $X$ vanish (including $\pi_0(X,\ast)$, so $X$ is pathconnected). If $X$ is a $K(G,1)$-space and $Y$ is a $K(H,1)$-space, then, assuming both spaces are homotopy equivalent to CW complexes, we have a bijection $\phi: [(X,\ast),(Y,\ast)] \to \hom(G,H)$ of homotopy classes of base point preserving maps and group homomorphisms, where $\phi([f]) = \pi_1(f)$. $\endgroup$ – Alexander Thumm May 30 '12 at 10:06
  • $\begingroup$ Now the torus is a $K(\mathbb Z^2,1)$-space, since $\pi_1(T^2, \ast) = \mathbb Z^2$ and all other homotopy groups vanish (the universal cover of $T^2$ is $\mathbb R^2$, which is contractible). $\endgroup$ – Alexander Thumm May 30 '12 at 10:06
  • $\begingroup$ The proof you are asking for then becomes a special case of the injectivity of $\phi: [(T^2,\ast), (T^2,\ast)] \to \hom(\mathbb Z^2, \mathbb Z^2) $ as defined above. In order to understand such a proof, you should be familiar with basic definitions of CW complexes and the cellular approximation theorem. $\endgroup$ – Alexander Thumm May 30 '12 at 10:15

I will give an elementary proof of the problem using the fact, that $T^2$ is a topological group and that its universal cover is contractible. We will start with some constructions in homotopy theory of topological groups, which are required to understand the proof given below and added here for convenience.

First note, that $\mathbb R^2$ forms a topological group under addition: $(x,y) + (x',y') := (x+x',y+y')$ and that $\mathbb Z^2$ is a discrete normal subgroup thereof. We identify $T^2$ with the quotient of $\mathbb R^2$ by $\mathbb Z^2$, so that $T^2$ again becomes a topological group under addition. Moreover the quotient map $p: \mathbb R^2 \to T^2$ becomes a group homomorphism and is easily seen to be the universal covering projection ($\mathbb Z^2$ discrete subgroup $\Rightarrow$ $p$ is covering projection; $\mathbb R^2$ is contractible $\Rightarrow$ $p$ is universal).

Next, we observe, that for $[f],[g]\in [(X,\ast),(T^2,0)]$ the sum $[f]+[g] := [f+g]$ is well defined, turning $[(X,\ast),(T^2,0)]$ into a group. The same arguments show that $[(X,\ast),(\mathbb R^2,0)]$ is a group under point wise addition of representatives as well (as is $[(X,\ast),(G,1)]$ for any topological group $G$ with unit $1$), and that the map $p_\sharp : [(X,\ast),(\mathbb R^2,0)] \to [(X,\ast),(T^2,0)]$ given by $p_\sharp([f]) := [p \circ f]$ is a group homomorphism.

Now $\pi_1(T^2,0) = [(S^1,1), (T^2,0)]$ is a group in two ways, by means of composition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha \ast \beta]$ and by means of point wise addition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha + \beta]$.

Both operations share the same unit, the (class of the) constant loop sending everything to $0 \in T^2$ and denoted simply by $0: (S^1,1) \to (T^2,0)$. We can also observe, that for any loops $\alpha, \beta, \gamma, \delta$ we have $(\alpha + \beta) \ast (\gamma + \delta) = (\alpha + \gamma) \ast (\beta + \delta)$. Therefore $$[\alpha]+[\beta] = ([\alpha] \ast [0]) + ([0] \ast [\beta]) = ([\alpha] + [0]) \ast ([0] + [\beta]) = [\alpha] \ast [\beta],$$ hence the two operations are in fact the same on $\pi_1(T_2,0)$. The same argument can be used to show the analogous statement for $\pi_1(\mathbb R^2,0)$ (or $\pi_1(G,1)$ for any topological group $G$ with unit $1$).

Now back to the problem:

Given two maps $\varphi, \psi: T^2 \to T^2$, such that for some point $x \in T^2$, we have $\varphi(x) = \psi(x) = x$ and $\pi_1(\varphi) = \pi_1(\psi): \pi_1(T^2,x) \to \pi_1(T^2,x)$, we want to show $\varphi \simeq \psi$, where the homotopy can be taken relative to $x$. Replacing $\varphi$ with $\xi \mapsto \varphi(\xi + x) - x$ and $\psi$ with $\xi \mapsto \psi(\xi + x) - x$ if necessary, we may assume $x=0$. It will then suffice to show, that $\chi \simeq 0$, where $\chi := \varphi - \psi$.

Since the induced map $\pi_1(\chi): \pi_1(T^2,0) \to \pi_1(T^2,0)$ on fundamental groups is trivial (this is where we need all the constructions for topological groups), we can lift $\chi$ to a map $\bar{\chi}: (T^2,0) \to (\mathbb R^2,0)$ with $\chi = p \circ \bar{\chi}$.

We now define $H: T^2 \times I \to T^2$ by $H(x,t) = p(t\bar\chi(x))$, which is easily checked to be the required homotopy $0 \simeq \chi$.

  • $\begingroup$ Note that this prove applies to any topological group $G$ with unit $1$ with contractible universal cover and any pair of maps $\varphi,\psi: X \to G$ with $\pi_1(\varphi) = \pi_1(\psi): \pi_1(X,x_0) \to \pi_1(G,1)$. $\endgroup$ – Alexander Thumm May 30 '12 at 12:53

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