I am starting to study fibre bundles and I came across the following.
If $(E,p,B,F)$ is a locally trivial fibration, and $g:B'\rightarrow B$ a map (continuous), then define $$g^\#E= \{(x,b') : p(x)=g(b')\}$$ $$p':g^\#E\rightarrow B' $$ $$p'(x,b')=b' $$
Then, apparently, it is easy to see that $(g^\#E, p', B',F) $ is a locally trivial fibration.
I tried showing this using the definition:
For each element $b'\in B'$ we have $g(b')\in B$. Since we have that $(E,p,B,F)$ is a locally trivial fibration, there must be an open set $U$ (a neighborhood of $g(b')$) such that 
is a product bundle.
So there is a homeomorphism $$\phi:U\times F \rightarrow p^{-1}(U)\subset E$$ such that $p\phi(u,y)=u$. Using this I defined a candidate for homeomorphism $$\phi':V\times F \rightarrow p'^{-1}(V)\subset g^\#E\subset E\times B' $$ $$\phi'(v,x) = (\phi(g(v),x), v)$$ in which $V=g^{-1}(U)\subset B'$.
I was able to show that $\phi'$ is a well-defined continuous bijection.
So I only need to show it is an open map (or a closed map) to conclude it is a homeomorphism, but I wasn't able to do so until now. Any ideas?
There might be a simpler way to prove this since the book mentioned it can be easily seen.
EDIT(SOLVED IT): I was able to show that the map above is open. I used the fact that $$\phi'(V_o\times F_o) = (\phi(U\times F_o)\times V_o)\cap p'^{-1}(V)$$