Let $X,Y$ be Hausdorff and $f,g : X \to Y$ continuous. Let $A \subset X$ be dense and suppose that $f\mid_A = g\mid_A$. Show that $f=g$.

Let $X,Y$ be Hausdorff and $f,g : X \to Y$ continuous. Let $A \subset X$ be dense and suppose that $f\mid_A = g\mid_A$. Show that $f=g$.

Suppose the contrary that $f \ne g$. This implies that there exists $x \in X$ such that $f(x)\ne g(x)$. As $Y$ is Hausdorff we can consider neighborhoods $U_{f(x)}$ and $U_{g(x)}$ for which we have that $U_{f(x)} \cap U_{g(x)} = \emptyset$. Now $$f^{-1}(U_{f(x)} \cap U_{g(x)})=f^{-1}(U_{f(x)}) \cap f^{-1}(U_{g(x)})=\emptyset$$ but this is a contradiction since both of them contain at least $x$?

I'm not sure if the proof is correct since I'm not using the density of $A$ at all. Isn't it true that $x \in f^{-1}(U_{f(x)}) \cap f^{-1}(U_{g(x)})$?

Solution 1:

Why would $f^{-1}[U_{g(x)}]$ contain $x$?

Without density of $A$, the claim is obviously false. For example, consider $A=\emptyset$.

Notice that the density of $A$ means that there is a net in $A$ convergent to $x$.

(You also only need $Y$ to be Hausdorff, $X$ can be arbitrary, as far as I can tell.)

Solution 2:

Consider $f^{-1}[U_{f(x)}] \cap g^{-1}[U_{g(x)}]$ which is open in $X$ by continuity of $f$ and $g$, and non-empty as $x$ is in it.

So it intersects the dense set $A$ in some $a$. For $a\in A$ we know $f(a)=g(a)$ and $f(a) \in U_{f(x)}$ and $g(a) \in U_{g(x)}$ which together contradicts the disjointness of the chosen neighbourhoods in $Y$. $X$ being Hausdorff is superfluous.