I apologize if the subject doesn't accurately describe my question.
Let $F_2$ denote the free group on two generators.
Suppose you have some group homomorphism $A : \mathbb{Z}^2\rightarrow\mathbb{Z}^2$. There are many ways to lift $A$ to a map $\alpha : F_2\rightarrow F_2$ (ie, the square diagram with vertical maps being abelianizations, commutes).
Firstly, am I right in saying that any two lifts of $A$ to an endomorphism of $F_2$ differ by some inner automorphism of $F_2$?
Now let $\alpha : F_2\rightarrow F_2$ denote a particular lift of the map on $\mathbb{Z}^2$ sending $(x,y)\mapsto(mx,my)$ for some positive integer $m$.
Let $E$ be a complex elliptic curve (ie, a complex torus). Let $[m] : E\rightarrow E$ denote the multiplication by $m$ map. Let $E^*$ denote the same curve, with the identity point removed. We identify $F_2$ with $\pi_1(E^*)$, and $\mathbb{Z}^2$ with $\pi_1(E)$, with suitable base points. The inclusion maps $E^*\hookrightarrow E$ on the level of fundamental groups become exactly abelianization maps $F_2\rightarrow \mathbb{Z}^2$. The map $[m]$ gives some map $\mathbb{Z}^2\rightarrow\mathbb{Z}^2$, so we're in a similar situation to what I described earlier.
Since the universal cover of $E,E^*$ are $\mathbb{C},\mathcal{H}$(the upper half plane) respectively, both are Eilenberg Maclane spaces. By a theorem described in Hatcher, if $Y$ is an Eilenberg-Maclane space, and $X$ a CW complex, then for homomorphism $\pi_1(X,x_0)\rightarrow\pi_1(Y,y_0)$, there is a map $X\rightarrow Y$ sending $x_0$ to $y_0$ which induces the given map on fundamental groups. Furthermore, this map is unique up to homotopy fixing $x_0$.
My questions are as follows:
- Is the theorem true for $X = E^*$? (ie, is it true that $E^*$ can be realized as a CW-complex or perhaps the theorem can be generalized to include the case $X = E^*$?)
- Assuming the answer to (1) is yes, there should be a map $E^*\rightarrow E^*$ inducing $\alpha$ on fundamental groups. How can we best describe this map? This map certainly cannot be a restriction/lift of $[m]$, since every $m$-torsion point would want to map to the identity of $E$, which is not in $E^*$. On the other hand, this map is "unique", which generally says it should somehow be close to a restriction/lift of $[m]$ to $E^*$...
Any help would be appreciated.
Thanks,
Will
No, using a free basis $F_2 = \langle a,b \rangle$, the map given by $f(a)=a$, $f(b) = a b^3 a^3 b^{-1} a^{-4} b^{-1}$ is not an inner automorphism but it induces the identity on $\mathbb{Z}^2$, as does the identity automorphism $g(a)=a$, $g(b)=b$.
The punctured torus $E^*$ deformation retracts to a 1-dimensional complex $R_2$, the "rank 2 rose" consisting of two loops touching at a point, namely the union of an $a$-curve and a $b$-curve in $E^*$. So yes, $E^*$ is an Eilenberg-Maclane space, because it is homotopy equivalent to the 1-complex $R_2$.
There are many such maps, but one has no expectation that they are anything other than yucky. The lifted endomorphisms of the map $[m]$ are many and varied. For instance $f(a)=a^m$, $f(b)=b^m$ is one, but also $g(a)=a^m$, $g(b) = b a^{42} b a^{71} b^{m-2} a^{-113}$. If you represent each of these formulas as a topological map on $R_2$ then you get a map on $E^*$ as a composition $$E^* \to R_2 \to R_2 \to E^* $$ where the first map is the deformation retraction, the second is the formula $f$ or $g$, and the third is the inclusion. Yuck.
Quite the opposite. Whatever this map might be, as I have tried to demonstrate it is very far from "unique".