Showing that a function is continuous

Question

Let $A$ be a topological space, and let $B$ be a quotient space of it. I have defined a continuous function $F:A\times[0,1]\to B$ such that it factors into a function $G:B\times[0,1]\to B$, i.e. that $F$ is the same as projecting the first argument to $B$ and then applying $G$ to the arguments. Is there now any way to show that $G$ is continuous, given that I know that $F$ is continuous?

Answer 1

So the point is that since $B$ is a quotient of $A$, then $B\times [0,1]$ is a quotient of $A\times [0,1]$.

Thus, if we let $\phi: A\times [0,1] \to B\times[0,1]$ denote the quotient map (in terms of the notation in the comments $\phi = (\gamma, id_{[0,1]})$), then we have

$F = G\circ \phi$.

Thus, since $F$ is continuous, so is $G$ by the definition of the quotient topology on $B\times[0,1]$.