Group theory problem 2

[Obelix]

Ok, we've warmed up enough! Now for the real thing:

 

Let [math](G,\cdot,T)[/math] and [math](H,\ast,S)[/math] be topological groups, [math]H[/math] being connected as a topological space, and let [math]f : G \rightarrow H: g \rightarrow f(g)[/math] be a homomorphism of topological groups, i.e.: [math]f : G \rightarrow H :[/math] Continuous, and: [math]f(g_1\cdot g_2)= f(g_1) \ast f(g_2)[/math] (or [math]f(g_1 g_2) = f(g_1) f(g_2)[/math], in simplified notation) [math]\forall g_1, g_2 \in G[/math]. Suppose moreover that [math]int_{S}f(G) \equiv f(G)^{\circ} \neq \emptyset[/math], i.e.: [math]\exists \, g \in G, U \in S[/math] such that: [math]f(g) \in U \subseteq f(G)[/math]. Then [math]f(G) = H[/math], i.e.: [math]f : G \rightarrow H[/math] is onto.

 

Can anyone suggest a proof? Then I will explain why I picked this problem up.

Edited May 23, 2011 by Obelix

[DrRocket]

Ok, we've warmed up enough! Now for the real thing:

 

Let [math](G,\cdot,T)[/math] and [math](H,\ast,S)[/math] be topological groups, [math]H[/math] being connected as a topological space, and let [math]f : G \rightarrow H: g \rightarrow f(g)[/math] be a homomorphism of topological groups, i.e.: [math]f : G \rightarrow H :[/math] Continuous, and: [math]f(g_1\cdot g_2)= f(g_1) \ast f(g_2)[/math] (or [math]f(g_1 g_2) = f(g_1) f(g_2)[/math], in simplified notation) [math]\forall g_1, g_2 \in G[/math]. Suppose moreover that [math]int_{S}f(G) \equiv f(G)^{\circ} \neq \emptyset[/math], i.e.: [math]\exists \, g \in G, U \in S[/math] such that: [math]f(g) \in U \subseteq f(G)[/math]. Then [math]f(G) = H[/math], i.e.: [math]f : G \rightarrow H[/math] is onto.

 

Can anyone suggest a proof? Then I will explain why I picked this problem up.

 

This is rather obvious.

 

Since you have a predilection for posting trivial problems and have yet to produce anything resembling an elegant or insightful argument, you have an opportunity to redeem yourself. Produce one -- say six lines or less.

 

No cheating by copying a solution from any of the other forums in which you have posted the same problem.

Edited May 23, 2011 by DrRocket

[mooeypoo]

!

Moderator Note

Moved to homework help.

Obelix, we'll be happy to help, but we're not here to feed you answers with a spoon. Please show us some of your suggested work, something you started trying or where you got stuck, and we will be happy to help you proceed.

And please post your questions where they SHOULD be posted - in homework help section. The other forums are for debates, not direct problem-solving.

[DrRocket]

Ok, we've warmed up enough! Now for the real thing:

 

Let [math](G,\cdot,T)[/math] and [math](H,\ast,S)[/math] be topological groups, [math]H[/math] being connected as a topological space, and let [math]f : G \rightarrow H: g \rightarrow f(g)[/math] be a homomorphism of topological groups, i.e.: [math]f : G \rightarrow H :[/math] Continuous, and: [math]f(g_1\cdot g_2)= f(g_1) \ast f(g_2)[/math] (or [math]f(g_1 g_2) = f(g_1) f(g_2)[/math], in simplified notation) [math]\forall g_1, g_2 \in G[/math]. Suppose moreover that [math]int_{S}f(G) \equiv f(G)^{\circ} \neq \emptyset[/math], i.e.: [math]\exists \, g \in G, U \in S[/math] such that: [math]f(g) \in U \subseteq f(G)[/math]. Then [math]f(G) = H[/math], i.e.: [math]f : G \rightarrow H[/math] is onto.

 

Can anyone suggest a proof? Then I will explain why I picked this problem up.

 

 

!

Moderator Note

Moved to homework help.

 

Obelix, we'll be happy to help, but we're not here to feed you answers with a spoon. Please show us some of your suggested work, something you started trying or where you got stuck, and we will be happy to help you proceed.

 

And please post your questions where they SHOULD be posted - in homework help section. The other forums are for debates, not direct problem-solving.

 

Obelix, we'll be happy to help, but we're not here to feed you answers with a spoon. Please show us some of your suggested work, something you started trying or where you got stuck, and we will be happy to help you proceed.

 

And please post your questions where they SHOULD be posted - in homework help section. The other forums are for debates, not direct problem-solving.[/modnote]

 

Since this problem is unrelated to earlier problems posted by Obelix, and since he has been silent for a couple of days (in at least two forums where this same problem was posted) I will assume that it is too late for a solution to be used in a manor contrary to academic ethics. Also, it is quite clear that Obelix cannot produce a proof himself.

 

So, for those who might find this problem a bit interesting here is a solution. It assumes some elementary facts regarding topological groups.

 

Proof: Since [math]G(T)[/math] has non-empty interior and since [math]G[/math] is continuous homomrphism we can, by translation if necessary, assume that [math]G(T)[/math] contains a neighborhood of the identity. Since [math]G[/math] it is a group homomorphism [math]G(T)[/math] contains the subgroup of [math]H[/math] generated by that neighborhood. But a connected group is generated by any neighborhood of the identity. QED

 

 

For anyone interested in pursuing the topic of topological groups I recommend the classic treatise by L.S. Pontryagin Topological Groups. For any electrical engineering control theory types, this is the Pontryagin of "the Pontryagin maximum principle". For any interested in harmonic analysis, this is the Pontryagin of "Pontryagin duality". For those interested in vector bundles this is the Pontryagin of "Pontryagin classes". He did rather well, as a mathematician at least (not necessarily as a nice guy), for someone who was blinded at age 14.