Let $f_1:X_1\rightarrow X$ and $f_2:X_2\rightarrow X$ be two maps from two topological spaces to a set X. We equip $X$ with the topology coinduced by $f_1$ and $f_2$, that is to say, a subset $U\subset X$ is open in $X$ iff $f_1^{-1}(U)$ is open in $X_1$ and $f_2^{-1}(U)$ is open in $X_2$.
Let Y be another topological space. Consider the space $X \times Y$ equipped with the product topology. Notice that we have two maps $F_1 = f_1 \times id_{Y}:X_1 \times Y \rightarrow X \times Y$ and $F_2 = f_2 \times id_{Y}:X_2 \times Y \rightarrow X \times Y$.
My question is: whether $F_1$ and $F_2$ coinduce the same topology with the product topology on $X\times Y$?
I know that the product topology on $X \times Y$ is less than the topology coinduced by $F_1$ and $F_2$.
They are not the same in general. Thanks to Thorgott's comment. I find a counterexample.
Let $X_1 = X = [0,1]$. Let $c:X_1 \rightarrow X$ defined by $c(x) = \frac{1}{2}$ for all $x\in[0,1]$. Then the coinduced topology on $X$ by $c$ is the discrete topology. Let $Y = [0,1]$. Consider the subspace$A = (\frac{1}{4},\frac{3}{4})\times(\frac{1}{4},\frac{3}{4}) \cup \{\frac{1}{4}\}\times [\frac{1}{4},\frac{3}{4}] \subset X \times Y = [0,1]^2$.
$A$ is not open in the product topology, otherwise the projection on Y would be open. But A is open in the topology coinduced by $c\times id_Y$ because $(c\times id_Y)^{-1}(A) = X_1 \times (\frac{1}{4},\frac{3}{4})$, which is open in $X_1 \times Y$.