Let $f : X \to Y $ be a function and suppose that $\mathcal S $ is a topology on X.
its states that $\mathcal T = \{ U \subset Y | f^{-1}(U) \in \mathcal S \}$ is a topology on Y. i think that in this case $f^{-1}(U)$ is the preimage of U not inverse. i think we need f to be more that just a function don't we need f to be onto or continous so we know that $ f^{-1} (Y) \in \mathcal S $ is it implicit from the question that f is one of these things? oddly the other two properties are easy w.o it.
finite intersection is clear if you take any $U_1, U_2 \in \mathcal T$ then we consider $f^{-1}(U_1) \cap f^{-1}(U_2) $ but we have that $f^{-1}(U_1), f^{-1}(U_2) $ are both in $ \mathcal S $ so its closed under finite intersection by induction so $f^{-1}(U_1) \cap f^{-1}(U_2) \in \mathcal S $ the argument for arbitrary intersection is similar.
its just the $\emptyset, Y $ im not sure how to find.
Few remarks to this question:
The quotient topology $\mathcal{T}$ defined in this way is the largest topology on $Y$ such that $f$ is $\mathcal{S}/\mathcal{T}$-continuous.
A similar notion is the initial topology on $X$. Given a topology $\mathcal{T}$ on $Y$, the initial topology on $X$, denoted by $\mathcal{S}$, is the smallest topology on $X$ such that $f$ is $\mathcal{S}/\mathcal{T}$-continuous. $\mathcal{S}$ is given explicitly by $\mathcal{S}=\{f^{-1}(A) \mid A\in\mathcal{T}\}$.
These two topologies are dual concept, sharing some kind of universal properties. For example, let $f:X\rightarrow Y$ be a map, $\mathcal{T}$ be a topology on $Y$, and $\mathcal{S}$ be the initial topology on $X$ induced by $(f,\mathcal{T})$. Then, for any topological space $(U,\mathcal{U})$ and any map $\theta:U\rightarrow X$, $f\circ\theta$ is $\mathcal{U}/\mathcal{T}$-continuous if and only if $\theta$ is $\mathcal{U}/\mathcal{S}$-continuous.
For the case of initial topology, the notion generalized to an arbitrary family of topological spaces. Let $X$ be a set. Let $I$ be a non-empty index set. For each $i\in I$, let $(Y_i,\mathcal{T_i})$ be a topological space and let $f_i:X\rightarrow Y_i$ be a map. We can always find a smallest topology on $X$, denoted by $\mathcal{S}$, such that for each $i\in I$, $f_i$ is $\mathcal{S}/\mathcal{T}_i$-continuous. Note that $\bigcup_{i\in I} \{f_i^{-1} (A)\mid A\in\mathcal{T}_i\}$ is a sub-base of $\mathcal{S}$.
Note that relative topology is a kind of initial topology. For, let $(Y,\mathcal{T})$ be a topological space and let $X$ be a subset of $Y$. Let $i:X\rightarrow Y$ be the inclusion map defined by $i(x)=x$. We can verify that the usual relative topology on $X$ is indeed the initial topology induced by $(i,\mathcal{T})$.
For Setting 4, we have an important example: Tychonoff Topology (or Product Topology). Let $I$ be a non-empty index set. For each $i\in I$, let $(X_i,\mathcal{T}_i)$ be a topological space. Let $X= \prod_{i\in I} X_i$. That is, $X = \{f\mid f:I\rightarrow \bigcup_{i\in I} X_i \mbox{ such that } f(i)\in X_i\}$. For each $i\in I$, let $\pi_i : X\rightarrow X_i$ be the canonical projection onto the $i$-th coordinate space $X_i$, that is $\pi_i(f) = f(i)$, $f\in X$. Then, the Tychonoff Topology on $X$ is the initial topology induced by the family $\{(\pi_i, \mathcal{T}_i)\mid i\in I\}$. Tychonoff Topology is important in analysis and it shares may nice properties. For example, it is compact iff $\mathcal{T_i}$ is compact for each $i\in I$ (the Tychonoff Theorem, whose proof requires the Axiom of Choice). Tychonoff Theorem has a lot of important consequences in analysis. For example, for a Banach space $X$, the closed unit ball of the dual space $X^\ast$ is weak-$\ast$ compact (Banach-Alaoglu Theorem).