Ontoness and Induced Maps on Fundamental Group.

[Bacle]

Hi, everyone:

Given a top space X, and a homeo. h: X--->X , we get an induced map

(by functoriality ) h_*: Pi_1(X)---> Pi_1(X) . We can also write

the map as a map g: Aut(X) --->Hom(Pi_1(X),Pi_1(X))

Is the map g always surjective.? . Almost definitely not injective, since

the Fund. Group functor is constant on homotopic maps, but I have no idea

how to tell if it is injective.

I have no idea. Anyone know.?

Thanks

 

[Tinyboss]

If f is a continuous map between topological spaces, then it induces a homomorphism f* of their fundamental groups. If f has a continuous inverse, then (f^{-1})_*=(f_*)^{-1}. (This is because star distributes over function composition.) From that you get that homeomorphisms of topological spaces induce isomorphisms of fundamental groups.

 

[quasar987]

The OP is asking whether given any permutation of the fundamental group there is a self-homeomorphism of the space that induces this permutation.

 

[Tinyboss]

Oops, I did totally misread that. I'd say the answer is no. The fundamental group of the wedge product of a circle and a circular strip (circle cross an interval) is Z+Z, but we can't send a generator of one to a generator of the other via a self-homeomorphism.

 

[quasar987]

Good example!

 

[Tinyboss]

The example works but I botched the homotopy: the group should be the free product \mathbb{Z}*\mathbb{Z}...I had homology on the brain when I wrote that. But swapping generators is still an automorphism of the fundamental group that can't be induced by a self-homeomorphism of the space.

 

[Bacle]

Thanks to both:

I think the answer is yes in this specific case: One just has to find images for
a generating set for Z(+)Z , which is finitely-generated ---3 matrices are enough.

And, re the mapping class group, thanks for the Rolfsen source. The idea is
nice:

Consider automorphisms h: T^2-->T^2 . By functoriality, we get a map

Pi_1(T^2)-->Pi_1(T^2) , or , in a more general way, we get a homomorphism :

f: Aut(T^2) --->Aut(Z(+)Z) ; a homeo. on the left and a homomorph. on the right.

Then we know Aut(Z(+)Z) ~ Gl(2,Z) .

So, ultimately, we have a homomorphism :

h: Aut(T^2)--->Gl(2,Z)

Since we "have proved" ontoness ( by finding images for the generating set),

we just mod out by the kernel , which is the set of maps that are isotopic to

the identity. I think that gives us the mapping class group for the torus.

 

[quasar987]

What are your 3 generating matrices?