# measure theory – \$sigma\$-algebra generated by a function having as target a Borel space

Let $$X, Y$$ be sets, $$f:X rightarrow Y$$ be a function and suppose we have a set of subsets of $$Y$$, $$tau$$; unless the answer to my question requires it, I think $$tau$$ can be arbitrary, not necessarily a topology.

Let $$H : = sigma({ f^{-1}(U): U in tau })$$ and $$G := left{ f^{-1}(M): M in sigma(tau) right}$$. Is it true that $$H = G$$?

I can see that $$H subseteq G$$ because $${f^{-1}(M): Min sigma(tau) }$$ is a $$sigma$$-algebra by set theoretic stuff, which contains $${ f^{-1}(U): U in tau }$$.

But I can’t find an argument for the converse inclusion. If I take $$G’$$ another $$sigma$$-algebra containing $${ f^{-1}(U): U in tau }$$, can I show that $$G subseteq G’$$? Probably yes, exploiting that $$sigma(tau)$$ is smallest, but I can’t see how.