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Let $f:X\to Y$, $Y$ is Hausdorff, and $f$ is continuous. How to prove that $f$ is injective if $X$ is Hausdorff?

It is easy enough to show that $f$ injective implies $X$ Hausdorff, and I have been able to find examples where, if $f$ is not injective, $X$ is not Hausdorff. However, is it possible to prove the above statement in general? I cannot seem to work it out from the definitions.

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    $\begingroup$ You have substantially changed the question with your edits. In the original version you had the condition that $X$ has the initial topology. Was it only by mistake that you have omitted this condition? $\endgroup$ May 14 at 16:45
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    $\begingroup$ hi rolodex. to add to @MartinSleziak's comment, the statement as you currently have it is false. for any topological spaces $X$ and $Y$, and any point $y\in Y$, the constant map $X\to Y$ given by $x\mapsto y$ for all $x\in X$ is always continuous. on the other hand if $X$ has at least two elements this map is not injective $\endgroup$ May 14 at 21:38

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Notice that $f\colon X\rightarrow Y$ is open by definition.

$\Rightarrow$: Assume $X$ is a Hausdorff space. If $x,x'\in X$ with $x\neq x'$, then there are open subsets $U,U'\subseteq X$ with $x\in U$, $x'\in U'$ and $U\cap U'=\emptyset$. Because of the definition of the induced topology, there are open subsets $V,V'\subseteq Y$ with $U=f^{-1}(V)$ and $U'=f^{-1}(V')$. We have $f(x)\in f(U)=V$, $f(x')\in f(U')=V'$ and $f^{-1}(V\cap V')=f^{-1}(V)\cap f^{-1}(V')=U\cap U'=\emptyset$. If $f(x)=f(x')$, then it would be in $V\cap V'$, which gives a contradiction. Therefore $f$ is injective.

Notice that if $f$ is surjective, you could also conclude $f^{-1}(V\cap V')=\emptyset\Rightarrow V\cap V'=\emptyset$ and therefore $f(x)\neq f(x')$.

$\Leftarrow$: Assume $f$ is injective. If $x,x'\in X$ with $x\neq x'$, then $f(x)\neq f(x')$ and there are open subsets $V,V'\subset Y$ with $f(x)\in V$, $f(x')\in V'$ and $V\cap V'=\emptyset$. Therefore $x\in f^{-1}(V)$ and $x'\in f^{-1}(V')$ (which are both open in $X$ due to the definition of the induced topology). Since $f^{-1}(V)\cap f^{-1}(V')=f^{-1}(V\cap V')=f^{-1}(\emptyset)=\emptyset$, $X$ is a Hausdorff space.

EDIT 1: An analogy of the theorem does not hold for the coinduced topology. Let $f\colon(X,\mathcal{P}(X))\rightarrow Y$ then $\mathcal{O}_Y=\{V\subseteq Y|f^{-1}(V)\in\mathcal{O}_X=\mathcal{P}(X)\}=\mathcal{P}(Y)$. Both $X$ and $Y$ are Hausdorff, but $f$ does not need to be injective nor surjective.

The analogy of the other direction holds: If $X$ is Hausdorff and $f$ is injective, then $Y$ is Hausdorff.

EDIT 2: To prove this, we first notice that $f$ injective implies $f$ open. If $U\subseteq X$ is open, then $f^{-1}(f(U))=U\subseteq X$ is open, so by definition $f(U)\subseteq Y$ is open.

Let $y,y'\in Y$ with $y\neq y'$. We can assume $y,y'\in\operatorname{img}(f)$ since $Y\setminus\operatorname{img}(f)$ is discrete. Let $x,x'\in X$ with $x\neq x'$ be the respective preimages, then there are open subsets $U,U'\subseteq X$ with $x\in U$, $x'\in U'$ and $U\cap U'=\emptyset$. Since $f$ is open, $y\in f(U)$ and $y'\in f(U')$ are open neighborhoods. Since $f$ is injective, $f(U)\cap f(U')=f(U\cap U')=f(\emptyset)=\emptyset$.

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