When discussing with awllower about [this question](https://math.stackexchange.com/questions/47359/if-a-subgroup-of-an-algebraic-group-is-solvable-is-its-closure-necessarily-solv), I begin to think about another one:

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> For a topological group $G$ and a subgroup $H$, is it true that $[\overline{H}, \overline{H}] = \overline{[H,H]}$? 
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where $[H,H]$ denote the derived subgroups of $H$.

I think, if I define the map $\phi: G \times G \rightarrow G, (x,y) \mapsto xyx^{-1}y^{-1}$, then the question will become the equality between $\phi(\overline{H}\times \overline{H})$ and $\overline{\phi(H \times H)}$. Giving $G \times G$ the product topology, we will have $\phi(\overline{H} \times \overline{H}) \supseteq \overline{\phi(H \times H)}$ if $\phi$ is closed; and $\phi(\overline{H} \times \overline{H}) \subseteq \overline{\phi(H \times H)}$ because $\phi$ is continuous. (I hope I am not mistaken in thinking that $\overline{H \times H} = \overline{H} \times \overline{H}$, which is a condition of the above.) 

So, a derived question is:

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> For a topological group $G$, is the map $\phi: G \times G \rightarrow G, (x,y) \mapsto xyx^{-1}y^{-1}$ closed if $G \times G$ is given the product topology?
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For the case when $G$ is an algebraic group, the topology will be changed to Zariski topology. Then is $\phi$ a closed map? Does $\phi(\overline{H}\times \overline{H})$ equal $\overline{\phi(H \times H)}$?

In fact I am more concerned with the algebraic group case. But if the first case is dealt, hopefully the algebraic case will become easier.

Thanks to everyone.