Is the following statement true?
Let $f:X\rightarrow Y$ be a function and $\mathcal{Z}\subset Y$ be the subset generating the sigma algebra $\mathcal{A}$. Then $f^{-1}A:=\{f^{-1}A:A\in\mathcal{A}\}$ is the smallest sigma algebra generated by $f^{-1}\mathcal{Z}:=\{f^{-1}Z:Z\in\mathcal{Z}\}$
My attempt is as follows:
By contradiction, say $\mathcal{A}'$ is the smallest sigma algebra generated by $f^{-1}\mathcal{Z}$
$$f^{-1}\mathcal{Z}\subset\mathcal{A}'$$ $$\mathcal{A}'\quad \sigma\text{-algebra in } X \implies\{F\subset Y:f^{-1}F\in\mathcal{A}'\}\text{ is a sigma algebra in Y}$$
Now $\mathcal{Z}\subset \{F\subset Y:f^{-1}F\in\mathcal{A}'\}$ and $\mathcal{Z}\subset\mathcal{A}$.
This implies that $\mathcal{A}\subset \{F\subset Y:f^{-1}F\in\mathcal{A}'\}$ since $\mathcal{A}$ is the smallest sigma algebra containing $\mathcal{Z}$
I got stuck here. Am I on the right track? And is the statement true in the first place?
Yes you are on the right track.
So you have $\mathcal{A}\subset \{F\subset Y:f^{-1}F\in\mathcal{A}'\}$. This means that $f^{-1}\mathcal A\subset\mathcal A'$.
Conversely, since $\mathcal Z\subset\mathcal A$, we clearly have $f^{-1}\mathcal Z\subset f^{-1}\mathcal A$. Since $f^{-1}\mathcal A$ is a sigma algebra and $\mathcal A'$ is the smallest sigma algebra which contains $f^{-1}\mathcal Z$, we have $\mathcal A'\subset f^{-1}\mathcal A$.
You just showed that $f^{-1}\mathcal A\subset\mathcal A'\subset f^{-1}\mathcal A$. Therefore, $f^{-1}\mathcal A=\mathcal A'$.
The statement is true.
PS : There is no need to say "By contradiction" in the beginning since the proof is "direct".