Show that if $(X,*,\cal T)$ is a t. g. then $\text{cl}Y_1*\text{cl}Y_2=\text{cl}(Y_1*Y_2)$ and $(\text{cl}Y)^{-1}=\text{cl}Y^{-1}$ where $Y_1,Y_2,Y⊆X$ 
Definition
A topological group is a group $(X,*)$ equipped with a topology $\cal T$ with resepct the functions
$$
p:X\times X\ni (x_1,x_2)\longrightarrow x_1*x_2\in X\quad\text{and}\quad s:X\ni x\longrightarrow x^{-1}\in X
$$
are continuous.

Now for any $Y_1,Y_2,Y\in\mathcal P(X)$ we put
$$
Y_1*Y_2:=\{x\in X:x=y_1*y_2\text{ }\text{where }y_i\in Y_i\text{ for }i=1,2\}\quad\text{and}\quad Y^{-1}:=\{x\in X:x=y^{-1}\text{ for }y\in Y\} 
$$
so that we let to prove that the identities
$$
\operatorname{cl}(Y_1*Y_2)=\operatorname{cl}Y_1*\operatorname{cl}Y_2\quad\text{and}\quad\operatorname{cl}Y^{-1}=(\operatorname{cl}Y)^{-1}
$$
holds for any $Y_1,Y_2,Y\in\mathcal P(X)$. So if $A_{x_1*x_2}$ is an open neighborhood of $x_1*x_2\in\operatorname{cl}Y_1*\operatorname{cl}Y_2$ then $p^{-1}[A_{x_1*x_2}]$ is an open neighborhood of $(x_1,x_2)$ in $X\times X$ but it is a well know result that
$$
\operatorname{cl}(Y_1\times Y_2)=\operatorname{cl}Y_1\times\operatorname{cl}Y_2
$$
so that if $(x_1,x_2)\in\operatorname{cl}Y_1\times\operatorname{cl}Y_2$ then
$$
p^{-1}[A_{x_1*x_2}]\cap (Y_1\times Y_2)\neq\emptyset
$$
and so by the inclusion
$$
p\Big[p^{-1}[A_{x_1*x_2}]\cap(Y_1\times Y_2)\Big]\subseteq p\big[p^{-1}[A_{x_1*x_2}]\big]\cap p[Y_1\times Y_2]\subseteq A_{x_1*x_2}\cap(Y_1*Y_2)
$$
we conclude that
$$
A_{x_1*x_2}\cap(Y_1*Y_2)\neq\emptyset
$$
which implies $x_1*x_2\in\operatorname{cl}(Y_1*Y_2)$ and so finally
$$
\operatorname{cl}Y_1*\operatorname{cl}Y_2\subseteq\operatorname{cl}(Y_1*Y_2)
$$
Another possible way to infer the last inclusion is the follow: so by continuity of $p$ we observe that the inclusion
$$
Y_1\times Y_2\subseteq p^{-1}[Y_1*Y_2]\subseteq\operatorname{cl}p^{-1}[Y_1*Y_2]\subseteq p^{-1}\big[\operatorname{cl}(Y_1*Y_2)\big]
$$
holds but by continuity of $p$ the set $p^{-1}\big[\operatorname{cl}(Y_1*Y_2)\big]$ is closed so that effectively the inclusion
$$
\operatorname{cl}Y_1\times\operatorname{cl}Y_2=\operatorname{cl}(Y_1\times Y_2)\subseteq p^{-1}\big[\operatorname{cl}(Y_1*Y_2)\big]
$$
holds and thus finally we conclude that
$$
\operatorname{cl}Y_1*\operatorname{cl}Y_2=p[\operatorname{cl}Y_1\times\operatorname{cl}Y_2]\subseteq p\Big[p^{-1}\big[\operatorname{cl}(Y_1*Y_2)\big]\Big]\subseteq\operatorname{cl}(Y_1*Y_2)
$$
as we desidred.
So as you can see I am not able to prove that the inclusion
$$
\operatorname{cl}(Y_1*Y_2)\subseteq\operatorname{cl}Y_1*\operatorname{cl}Y_2
$$
holds so that I ask to prove it. Moreover is the inclusion
$$
\operatorname{cl}Y_1*\operatorname{cl}Y_2\subseteq\operatorname{cl}(Y_1*Y_2)
$$
well proved? Finally how prove that the identity
$$
\operatorname{cl}Y^{-1}=(\operatorname{cl}Y)^{-1}
$$
holds? So could someone help me, please?
 A: Both proofs of $\newcommand{\cl}{\operatorname{cl}}$
$$
\cl (Y_1) * \cl (Y_2) \subseteq \cl(Y_1 * Y_2) \tag1
$$
are correct. Here is another one:
Exercise: If $f \colon T \to S$ is continuous map between topological spaces, then for any $A \subseteq T$ one has that
$$
f[\cl(A)] \subseteq \cl(f[A]). \tag2
$$
(In fact, the converse also holds.)
Thus, $(1)$ follows from this with $f = p$ and $A = Y_1 \times Y_2$.

And, unless I missing something obvious, the other inclusion does not holds in general:
It is clear that $(\Bbb R,+)$ is a topological group with the usual topology, that $\Bbb Z$ and $\sqrt2\Bbb Z$ are closed, and $\Bbb Z+\sqrt2\Bbb Z$ is dense. Hence,
$$
\cl(\Bbb Z+\sqrt2\Bbb Z) = \Bbb R \neq \Bbb Z+\sqrt2\Bbb Z = \cl(\Bbb Z)+\cl(\sqrt2\Bbb Z).
$$

Finally, since $s$ is its own inverse, it is a homeomorphism; so, the above exercise tells us that
$$
s[\cl(Y)] = \cl(s[Y]).
$$
(If $f \colon S \to T$ is a homeomorphism, then by the exercise we have that $f^{-1}[\cl(f[A])] \subseteq \cl(f^{-1}[f[A]]) = \cl(A)$, and applying $f[\_]$ we obtain the equality in $(2)$.)
