An elliptic curve $E$ over $\Bbb{C}$ is defined over a subfield $⊆ℂ$ if there exists an elliptic curve $E_0$ given by a Weierstrass equation with coefficients in $$ such that $\newcommand{\Spec}{\operatorname{Spec}} E\cong E_0×_{\Spec(K)}{\Spec{\Bbb{C}}}$ as curves over $\Bbb{C}$.
Let $\newcommand{\Ell}{\operatorname{\mathcal{ELL}}} \Ell(\overline{\Bbb{Q}})$ be the quotient set $$ \frac{\{\text{elliptic curves over $\Bbb{C}$ which are defined over $\overline{\Bbb{Q}}$}\}}{(\text{isomorphisms over $\overline{\Bbb{Q}}$})}. $$ Let $\Ell( \Bbb{C})$ be the quotient set $$ \frac{\{\text{elliptic curves over $\Bbb{C}$ which are defined over $\overline{\Bbb{Q}}$}\}}{(\text{isomorphisms over $\Bbb{C}$})}. $$
Then, what is an natural map from $\Ell(\overline{\Bbb{Q}})$ to $\Ell( \Bbb{C})$ ? I think I need to use some kind of universal property of fiber product.
From the commments: If we choose an embedding $\newcommand{\Q}{\mathbb{Q}} \newcommand{\C}{\mathbb{C}} \iota: \overline{\Q} \hookrightarrow \C$, then every $\overline{\Q}$-isomorphism is also a $\C$-isomorphism. So we can simply map the $\overline{\Q}$-isomorphism class of an elliptic curve $E$ to its $\C$-isomorphism class.
This can be made concrete by noting that, over an algebraically closed field $L$, the isomorphism class of an elliptic curve is completely determined by its $j$-invariant, and every $j$-invariant occurs. (See Silverman's The Arithmetic of Elliptic Curves, Proposition III.1.4(b) and (c), p. 45.) In other words the $j$-invariant map \begin{align*} \newcommand{\Ell}{\operatorname{\mathcal{ELL}}} \Ell(L) &\overset{\sim}{\to} L\\ E &\mapsto j(E) \end{align*} is an isomorphism, and we have the following commutative diagram. $$ \require{AMScd} \begin{CD} \Ell(\overline{\Q}) @>{}>> \Ell(\C);\\ @V{j}V{}V @VV{j}V \\ \overline{\Q} @>>{\iota}> \C \end{CD} $$