Let $f:Y \rightarrow Z$ is continuous, injective and $Z$ is a Hausdorff, then $Y$ is also Hausdorff. Let $f:Y \rightarrow Z$ is continuous, 1-1 and $Z$ is a Hausdorff, I want to show $Y$ is also Hausdorff.
The followings are my trials :
Let $y_1, y_2 \in Y$ with $y_1 \neq y_2$ since $f$ is injective, images are differents, say $f(y_1) \neq f(y_2)$. And since $Z$ is Hausdorff we can conisder disjoint open neighborhoods of $f(y_1), f(y_2)$, denoted as $U_{1}, U_{2}$ respectively. Since $f$ is continuous, inverse image of open sets is again open. So $f^{-1}(U_{_1}), f^{-1}(U_2)$ are open in $Y$.
Now I want to show $f^{-1}(U_1) \cap f^{-1}(U_2) = \emptyset$.

The lecturer just says since $f$ is $1-1$ they are disjoint

, but I do not understand. Can anyone give me more detail about this process?
Actually I proved this via the properties of inverse images under intersection, $f^{-1}(U_1) \cap f^{-1}(U_2) = f^{-1}(U_1 \cap U_2) = f^{-1}(\emptyset) = \emptyset$ but I want to do in the above way. (want to understand what lecture says)
 A: In general $g^{-1}[A] \cap g^{-1}[B] = g^{-1}[A \cap B]$ as is easy to see:
$$x \in g^{-1}[A] \cap g^{-1}[B] \iff x \in g^{-1}[A] \land x \in g^{-1}[B] 
\iff g(x) \in A \land g(x) \in B \\
\iff g(x) \in A \cap B \iff x \in g^{-1}[A \cap B]$$
and this implies that if $V_1 \cap V_2 = \emptyset$ then also $g^{-1}[V_1] \cap g^{-1}[V_2] = \emptyset$  and this holds for any function, not just injective ones. We only need the 1-1 to ensure we can apply Hausdorffness to $f(x_1) \neq f(x_2)$.
A: The way you proved that they are disjoint is the "right way," in my view. In particular, for any function $g: A \to B$ and $V_1,V_2 \subseteq B$ such that $V_1 \cap V_2 = \emptyset$, we must have $g^{-1}(V_1) \cap g^{-1}(V_2) = \emptyset$. Otherwise we would send one element to two different places, which is essentially what you showed.
One-to-one functions do send disjoint sets to disjoint sets, so maybe that's what your lecturer was referring to, though it doesn't matter for this particular point. The injectivity is only needed in the first line of your proof.
A: Hint: if $f : Y \to Z$ is continuous and $1$-$1$, we can think of $Y$ as a subset of $Z$ such that every $Z$-open set is also $Y$-open, but $Y$ may have other open sets too. I.e., the topology of $Y$ is obtained from its topology viewed as a subset of $Z$ by adding $0$ or more extra open sets. Now think about the definition of the Hausdorff property: adding open sets can't invalidate that property.
