Extension of continuous map in topological space In the book _Simmons, George F._ , Introduction to topology and modern analysis, page no- 98, question no- 2, the problem is : **_Let $X$ be a topological space and a $Y$ be metric space and $f:A\subset X\rightarrow Y$ be a continuous map. Then $f$ can be extended in at most one way to a continuous mapping of $\bar{A}$ into $Y$._** I am trying to prove this way. Let $x_0\in \bar{A}-A$ and suppose that there is two extension $f$ and $g$ such that $f(x)=g(x)$ for $x\in A$. Now $f(x_0)\in \overline{f(A)}$ and $g(x_0)\in\overline {g(A)}$. So there exists a sequence $\\{f(x_n)\\}$ and $\\{g(y_n)\\}$ that converge to $f(x_0)$ and $g(x_0)$ respectively, where $x_n$ and $y_n$ belong to $A$ for all $n$. Then I am stuck!! Please help to complete the proof.

The following is a well-known result in point-set topology.

> **Proposition.** Two continuous functions $f, g \colon X \to Y$ from a topological space $X$ to a Hausdorff space $Y$, that coincide over a dense subset $D \subseteq X$, necessarly coincide everywhere.

_Proof._ Consider the set $$Z :=\\{x \in X \, | \, f(x)=g(x)\\} \subseteq X.$$ Then $Z$ is closed in $X$, since it is the preimage of the diagonal of $Y \times Y$ (that is closed because $Y$ is Hausdorff) via the continuous map $$h \colon X \to Y \times Y, \quad x \mapsto (f(x), \, g(x)).$$ On the other hand, by assumption $D \subseteq Z$ and so, since $D$ is dense in $X$, we obtain $$X = \bar{D} \subseteq \bar{Z} = Z,$$ that is $X = Z$ and the proof is complete.

Now we can get what you want from the Proposition above, because $A$ is dense in $\bar{A}$ and $Y$ is metric, hence Hausdorff.