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We have an epimorphism $f:G\to H$ between topological groups. We consider $\tau$ as the family of all images $f(U)$ where $U\subseteq G$ is open in $G$. The assertion is that $\tau$ is a topology over $H$ such that $(H,\tau )$ is a topological group. Then:

1) Clearly $\emptyset , H\in \tau$ and $\tau$ is closed under arbitrary union. If $U_1,U_2\subseteq G$ are open, then $f(U_1)\cap f(U_2)=f(U_1\cap NU_2)\in\tau$, where $N=Ker(f)$. Hence $\tau$ is a topology over $H$.

2) The function $i:H\to H$ defined by $i(h)=h^{-1}$ must be continuous on $(H,\tau)$. We take $f(U)\in \tau$ with $U\subseteq G$ open. Then $i^{-1}(f(U))=f(U^{-1})\in\tau$ because $f$ is a homomorphism.

3) But I can't prove that the function $m:H\times H\to H$ defined by $m(x,y)=xy$ is continuous yet. If $f(V)\in\tau$ with $V\subseteq G$ open, why is $m^{-1}(f(V))$ an open set in $H\times H$?

Thanks.

Talexius
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    Quick question: do you know that this is a topological group, but don't know how to prove it, or are you just conjecturing that it is one? – dfeuer Jul 16 '13 at 05:37
  • It must be a topological group. It is an assertion in a text book, but the autor didn't prove it. As I said, it only remains to prove that the multiplication is a continuous function. – Talexius Jul 16 '13 at 05:43

1 Answers1

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Pull it back to $G$.

Let $(h_1,h_2)\in H\times H$ be any point with the property $h_1h_2\in f(V)$. As $f$ is epi, there are points $g_1,g_2\in G$ such that $f(g_i)=h_i,i=1,2.$ Because $f$ is a homomorphism, $f(g_1g_2)=h_1h_2$. As $f^{-1}(f(V))=VN$, there exists an element $n\in N=\ker f$ such that $g_1g_2n\in V$. By replacing $g_2$ with $g_2n$ we can then assume that $g_1g_2\in V$.

Because the multiplication of $G$ is continuous, there exist open sets $V_1,V_2\subseteq G$ such that $g_1\in V_1, g_2\in V_2, V_1V_2\subseteq V$. It follows that for all $h_1'\in f(V_1), h_2'\in f(V_2)$ we have $$h_1'h_2'\in f(V_1)f(V_2)=f(V_1V_2)\subseteq f(V).$$ This means that $f(V_1)\times f(V_2)\subseteq m^{-1}(f(V)).$ As $(h_1,h_2)$ was arbitrary, it follows that $m^{-1}(f(V))$ is a union of open sets like $f(V_1)\times f(V_2)$, and is thus open by part 1.

Jyrki Lahtonen
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