# Does this collection of sets form a topology?

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.

## 2 Answers

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 $$C$$.

• (+1) I was trying to find such an example. Sep 9, 2021 at 9:47
• Thanks. I wasn't even considering the circle, as we haven't yet discussed the subspace topology. Sep 10, 2021 at 20:25

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!

• Yes it was. Thanks. (+1) Sep 10, 2021 at 20:27