Does this collection of sets form a topology?

Question

Let $X$ be a set, $(Y,\tau)$ a topological space, $f:X\rightarrow Y$ a surjection, and define $\tau_f=\{U\subset X\,|\,f(U)$ is $\tau$-open in $Y\}$. I'm trying to determine whether $\tau_f$ is a topology on $X$. It's not hard to see that $\varnothing,X\in\tau_f$. Moreover, since $$\bigcup_{\alpha\in\mathcal{J}} f(U_\alpha)=f\left(\bigcup_{\alpha\in\mathcal{J}} U_\alpha \right),$$ it's clear that $\tau_f$ is closed under arbitrary unions. But I'm having trouble showing $\tau_f$ is closed under finite intersections (an equality like the one above doesn't hold for intersections), and I could not find any counterexamples. I basically just looked at $X=Y=\mathbb{R}$ with different topologies on $Y$ (standard, lower-limit, discrete, indiscrete) and different choices of $f$ to try to find a counterexample, but failed.

I would appreciate a hint for the intersection part. Thanks in advance.

Answer 1

Try to map the real line onto unit circle: $f:\mathbb R\to C$, $f(x)=e^{ix}$. Take $A=[0,+\infty)$ and $B=(-\infty, 0]$. Take the usual topology on the unit circle. Now, $f(A)=f(B)=C$ - open, so in the new "topology" on $\mathbb R$, $A$ and $B$ would be "open". However, $f(A\cap B)=f(\{0\})=\{1\}$ which is not open in $C$.

Answer 2

Here is another interesting example: Consider $$ f : (\mathbb{Z},\tau_f) \to (\mathbb{N},\mathcal{O}), \quad n \mapsto |n|, $$ where $\mathcal{O}$ denotes the cofinite topology on $\mathbb{N} = \{0,1,2,\ldots\}$. Let $U_1 := \mathbb{N}$. We see that $U_1 \in \tau_f$, since $f(U_1) = \mathbb{N} \in \mathcal{O}$. Similarly, we see that $U_2 \in \tau_f$. However, $f(U_1\cap U_2) = f(\{0\}) = \{0\}$, which is not open in $\mathcal{O}$.

I hope this was helpful!