Criteria for Hausdorff 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.
 A: Since the question was changed, but I answered before and also with directions swapped: The original question was, if $Y$ had the induced topology, then $f$ is injective iff $X$ is Hausdorff. The backward direction does not use the induced topology on $Y$, which is already mentioned in the question.
$\Leftarrow$: 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 $f(x)\neq f(x')$.
$\Rightarrow$: Assume $f$ is injective. Notice also that $f\colon X\rightarrow Y$ is open by definition of the induced topology. 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, $f(U)\cap f(U')=f(U\cap U')=f(\emptyset)=\emptyset$.
A: As it was already mentioned in the question, the forward direction is easy, but I just had an idea for a very elegant proof, which I wanted to share and which I think deserves its own answer:
Since $f$ is continuous, $f\times f$ is continuous and preimages of closed subsets are closed. Since $f$ is injective, we have $\Delta_X=(f\times f)^{-1}(\Delta_Y)$. Since $Y$ is Hausdorff, $\Delta_Y$ is closed, therefore $\Delta_X$ is closed and $X$ is Hausdorff.
