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Let $X,Y$ be topological spaces, $\pi$ a surjective map $X\rightarrow Y$ such that $Y$ has the quotient topology induced by $\pi$. Let $I: [0,1]\rightarrow [0,1]$, $I(x)=x$.

Is it true that the quotient topology on $Y\times [0,1]$ given by the map $\pi \times I$ is the same as the product topology on $Y\times [0,1]$?

My try: $U$ is open in the quotient topology given by $\pi \times I \iff (\pi\times I)^{-1}(U) $ is open in $X \times [0,1]$, and now I'd like to use this to show that the product $A\times B$ of open sets $A$ in $Y$ and $B$ in $B$ form a basis for the quotient topology, but here I'm stuck.

Is it a good approach? Is my claim even true?

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