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Let $(G,\cdot,\mathscr{T})$ be a topological group, then $\cdot$ is indeed continuous, but is it open(close) mapping?

It is true at $(\mathbb R,+,\mathscr{T}_{Ord})$, so I guess it is also true in general.

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up vote 1 down vote accepted

It is an open mapping. It is not necessarily closed.

To see that it is an open mapping, define $m_z:G\to G$ by $m_z(g)=gz$. For each $z\in G$, the maps $m_z$ and $m_{z^{-1}}$ are continuous, because $m_z$ is a composition of maps $i_z:G\to G\times G$ given by $i_z(g)=(g,z)$ and $m:G\times G\to G$ given by $m(g,h)=gh$, i.e. the group operation. These maps are continuous, $m$ by definition of topological group, and $i_z$ because the preimage of $U\times V$, where $U$ and $V$ are open in $G$, is $$i_z^{-1}(U\times V)=\begin{cases}\emptyset;&z\notin V\\U;&z\in V\end{cases}$$ which is, in both cases, open. Note that this means that $m_z$ and $m_{z^{-1}}$ are inverse to each other. Since they are continuous inverses of each other, and continuous, they are homeomorphisms.

This implies that $m$ is open: if $U$ and $V$ are open in $G$, then $m(U\times V)=\bigcup_{y\in V}m_y(U)$, which is clearly open (since is suffices to verify continuity on a base).

To see that $m$ is not necessarily closed, we take $G=\mathbb R$ and $m(x,y)=x+y$. Define $A\subseteq\mathbb R^2$ by $A=\{(m,n\sqrt2)|\;m,n\in\mathbb Z\}$. It is not hard to see that this is a closed set. The image of this set under $m$ is $m(A)=\{m+n\sqrt 2|m,n\in\mathbb Z\}$, which is a countable dense set and thus cannot be closed.

Added: Here's a proof that $m(A)$ is indeed a countable dense set. Countability is clear, so we just have to prove that it is dense. To see this, let $1\geq\epsilon>0$. Then there exist $m_0,n_0\in\mathbb Z$ such that $0<m_0+n_0\sqrt2<\epsilon$, which we will now prove:

Suppose not. Let $S=\{r\in m(A)|\;r>0\}$. Then $\eta=\inf S$ is positive. Since $\sqrt2-1\in S$, we have that $\eta<1$. Let $k$ be the smallest integer such that $k\eta\geq1$. There are two cases: the first one is that $k\eta=1$.

In this case either there exist $m,n\in\mathbb Z$ such that $m+n\sqrt2=\eta$, which is clearly not the case (since $\sqrt2$ is irrational), or else there are no such $m,n\in\mathbb Z$. If there are no such $m,n\in\mathbb Z$, we can choose $m_1,n_1$ such that $\eta<m_1+n_1\sqrt2<2\eta$ and $m_2,n_2$ such that $\eta<m_2+n_2\sqrt2<m_1+n_1\sqrt2<2\eta$. Then $0<(m_1-m_2)+(n_1-n_2)\sqrt2<\eta$, which is a contradiction.

The second case is that $k\eta>1$. In this case, $(k-1)\eta<1$ and therefore $k\eta-1<\eta$. Let $\delta>0$ be a number such that $k\eta-1+k\delta<\eta$. Choose $m,n\in\mathbb Z$ such that $\eta<m+n\sqrt2<\eta+\delta$. Then $k\eta<km+kn\sqrt2<k\eta+k\delta$ and therefore $0<(km-1)+kn\sqrt2<k\eta-1+k\delta<\eta$. This is again a contradiction.

Therefore, indeed, for each $\epsilon>0$ there exist $m_0,n_0\in\mathbb Z$ such that $0<m_0+n_0\sqrt2<\epsilon$. Now let $(a,b)$ be an open interval in $\mathbb R$ and let $0<\epsilon<b-a$. Then there exists $k\in\mathbb Z$ such that $km_0+kn_0\sqrt2\in (a,b)$. This is because there is a $k\in\mathbb Z$ such that $a<k(m_0+n_0\sqrt2)$. Choose the smallest such $k$. (This exists because $m_0+n_0\sqrt2$ is positive.) It is impossible that $k(m_0+n_0\sqrt2)\geq b$, since otherwise $(k-1)(m_0+n_0\sqrt2)>a$ (because $m_0+n_0\sqrt2<b-a$), contradicting the minimality of $k$. This concludes the proof.

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That's all right. I have made an error for only considered the sum of two closed interval, whereas $+$ in $\mathbb R$ is exact not a close mapping. Thank you very much. – Popopo Nov 3 '12 at 11:11

$\cdot$ is open in general: Note that given an open $U \subseteq G^2$, we can write it as union of basic open sets $V_i \times W_i$, $V_i$, $W_i$ open in $G$. We have \begin{align*} \cdot(U) &= \cdot\left(\bigcup_i( V_i \times W_i)\right)\\ &= \bigcup_i (V_i \cdot W_i) \end{align*} And each $V_i \cdot W_i = \bigcup_{g\in V_i} gW_i$ is open as a union of open sets. Note, that $gW_i$ is open, as $g\cdot \colon G \to G$ is a homeomorphism.

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Right, thank you. – Popopo Nov 3 '12 at 10:49

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