How to prove or disprove the following statement on Hausdorff topology

Question

Let $X$ be a set and let $Y$ be a Hausdorff space. Let $f\colon X \to Y$ be a given mapping. Define $U \subset X $ to be open in $X$ if, and only if $U = f^{-1}(V)$ for some set $V$ open in $Y$. This defines a Hausdorff topology on $X$.

I have no idea how to start with.

Answer 1

There's nowhere else to start, but with definitions! That having been said, I'm assuming you know the definition of a topology.

Let $\tau$ be the collection of sets described in your question. That is, $\tau$ is the set of all $U\subseteq X$ such that $U=f^{-1}(V)$ for some open $V\subseteq Y$. Then,

$$X=f^{-1}(Y)$$

so $X\in\tau$.

$$\emptyset=f^{-1}(\emptyset)$$

so $\emptyset\in\tau$.

Let $U_{1},\ldots,U_{n}\in\tau$, say $U_{i}=f^{-1}(V_{i})$ where $V_{i}\subseteq Y$ is open. Then,

$$\bigcap_{1}^{n}U_{i}=\bigcap_{1}^{n}f^{-1}(V_{i})=f^{-1}\left(\bigcap_{1}^{n}V_{i}\right)$$

Certainly $\bigcap_{1}^{n}V_{i}$ is an open set of $Y$, so $\bigcap_{1}^{n}U_{i}$ is an element of $\tau$. Next we let $\{U_{\alpha}\}$ be a collection of elements of $\tau$, say $U_{\alpha}=f^{-1}(V_{\alpha})$ where $V_{\alpha}$ is open in $Y$. Then

$$\bigcup_{\alpha}U_{\alpha}=\bigcup_{\alpha}f^{-1}(V_{\alpha})=f^{-1}\left(\bigcup_{\alpha}V_{\alpha}\right)$$

Certainly $\bigcup_{\alpha}V_{\alpha}$ is open in $Y$, so $\bigcup_{\alpha}U_{\alpha}$ is an element of $\tau$. We then have that $\tau$ is a topology on $X$.

This topology is not necessarily Hausdorff

Let $X$ be the underlying set of $\mathbb{R}$ and let $Y=\{0\}$ with the indiscrete topology. Then $Y$ is Hausdorff, and the constant function $f:X\rightarrow Y$ can be our given map. The topology described above is then the indiscrete topology on $X$, which is not Hausdorff.

If the function is injective, then the topology is Hausdorff.

If $f$ is injective, then given distinct $x,y\in X$ we have that $f(x)$ and $f(y)$ are distinct in $Y$. Then, because $Y$ is Hausdorff there would be disjoint open sets $V_{1},V_{2}\subseteq Y$ containing $f(x)$ and $f(y),$ respectively. Then $f^{-1}(V_{1})$ and $f^{-1}(V_{2})$ are disjoint open sets of $X$ that contain $x$ and $y$, respectively. Thus $X$ equipped with $\tau$ is Hausdorff.

Answer 2

Let's suppose that $f$ is injective.

Let $x_1 \neq x_2 \in X$. Consider their images $f(x_1)$ and $f(x_2)$. They are different because $f$ is injective. Because $Y$ is Hausdorff, there exists two open subsets $U_1$ and $U_2$ of $Y$, with $f(x_1) \in U_1$ and $f(x_2) \in U_2$, and $U_1 \cap U_2 = \emptyset$.

Now $f^{-1}(U_1)$ and $f^{-1}(U_2)$ are open in $X$ by definition of open sets on $X$. They contain respectively $x_1$ and $x_2$. And they don't intersect, otherwise $U_1$ et $U_2$ would intersect.

This shows that $X$ is Hausdorff.

Answer 3

This is not true in general. Suppose $f(x)=f(y)$, then for all $U\subset X$ open with $x\in U$ there exists a $V\subset Y$ open such that $U=f^{-1}(V)$. Note that $x\in U$ if and only if $f(x)=f(y)\in V$ which implies that $y\in U$. So you need that $f$ is injective. If that holds you can follow the answer of TheSilverDoe.