Extending a domain of a function to closure of domain

Question

Let $X$ be a topological space , $Y$ be a metric space and $A$ be a subset of $X$. Let $f$ be a continuous function from $A$ to $Y$.

(Here $f$ is continuous in the sense that any inverse of image of an open set is open.)

How do I prove that there is at most one way to extend $f$ from $Cl(A)$ to $Y$ ?

( $Cl(A)$ is the intersection of all closed supersets of $A$ ).

In case of $\mathbb{R}$ we can verify this easily but I am not able to do the same for topological space. Any suggestion is appreciable.

Answer 1

Suppose that $f$ can be extended to two distinct continuous functions $F$ and $G$. Since $F\ne G$, there is some $a\in\overline A$ such that $F(a)\ne G(a)$. Let $\varepsilon=\frac12d\bigl(F(a),G(a)\bigr)$. Then$$B_\varepsilon\bigl(F(a)\bigr)\cap B_\varepsilon\bigl(G(a)\bigr)=\emptyset.$$Now, consider the set$$F^{-1}\left(B_\varepsilon\bigl(F(a)\bigr)\right)\cap G^{-1}\left(B_\varepsilon\bigl(G(a)\bigr)\right);\tag1$$it's an open subset of $X$ to which $a$ belongs. So, there is some $a^\star\in A$ which belongs to $(1)$ too. But this is impossible, since $F(a^\star)=f(a^\star)=G(a^\star)$, which implies that$$f(a^\star)\in B_\varepsilon\bigl(F(a)\bigr)\cap B_\varepsilon\bigl(G(a)\bigr).$$