If $X,Y$ are the Haudorff Space, can one deduce that , if $f$ is continous open mapping ,then $f$ is an injective?

Question

If $X,Y$ are the Haudorff Space, can one deduce that , if $f$ is a continous open mapping ,then $f$ is an injective?

Well, I think the condition Haudorff Space is necessary, for there is an example :

$$f:X\mapsto Y$$ Where $X,Y$ both have the trival topology, and $f$ is a surjective , obviously it is continous and an open mapping , but it is not injective.

Answer 1

Take $X$ to be $\mathbb{R}^2$ in the usual topology and $Y=\mathbb{R}$ in the usual topology and $f(x,y)=x$, the projection. Then $f$ is open, continuous and very non-injective, but onto. Both spaces are metric so certainly Hausdorff.

Projections are among the canonical examples of open and continuous maps. Many examples of this form exist.

Answer 2

Let $X$ be a space with $\lvert X \rvert > 1$ given the discrete topology. Singletons are open, so $X$ is Hausdorff. Let $x_0 \in X$.

Define $f : X \rightarrow X$ by $f(x) = x_0$ for $x \in X$.

$f$ is continuous and open, but neither injective nor surjective.