Prove that $\mathcal{G}$ is $\sigma$-algebra Let $X, Y$ be arbitrary spaces, $f: X \to Y$ arbitrary function. Prove that if $\mathcal{F}$ is $\sigma$-algebra of X then family
$$
\mathcal{G} = \{G \subset Y : f^{-1}(G) \in \mathcal{F}\}
$$
is a $\sigma$-algebra of Y.
First, I have proved that it is closed under complement:
For arbitrary $G \in \mathcal{G}$ we have
$$
G \in \mathcal{G} \implies f^{-1}(G) \in \mathcal{F} 
$$
then taking complement of that we notice it is also in $\mathcal{F}$ because $\mathcal{F}$ is $\sigma$-algebra.
$$
\left(f^{-1}(G)\right)^C \in \mathcal{F}
$$
At last, by logic laws
$$
f^{-1}(G^C) \in \mathcal{F}
$$
but that implies $G^C \in \mathcal{G}$ by definition of $\mathcal{G}$.
After that, I have proved that it is closed under contable sums. Taking countable number of sets
$$
 G_1, G_2, \cdots \text{ and } f^{-1}(G_1)\in \mathcal{F},  f^{-1}(G_2)\in \mathcal{F}, \cdots
$$
and because $\mathcal{F}$ is a $\sigma$-algebra, their sum
$$
 \bigcup\limits_{n=1}^{\infty} f^{-1}(G_n) \in \mathcal{F}
$$
that implies $\bigcup\limits_{n=1}^{\infty} G_n \in \mathcal{G} $.
I am stuck on third axiom - empty set has to be in $\mathcal{G}$. How can I tackle that?

The preimage of the empty set is also empty set! 
In fact, suppose $f(A)=\emptyset$ and $A\neq \emptyset$. There exists $x\in A$, therefore $y\in f(A)=\emptyset$, a contradiction.
