Let $\{(X_j,T_j\}_{j\in J}$ family of topological spaces and $\forall j\in J\ \mathcal R_j$ an equivalence relation on $X_j$. We condiderer now on $\sum_{j\in J}X_j$ the equivalence relation $\mathcal R$ given by: $$(x,j)\mathcal R (x',j')\iff x\mathcal R_j x',\ j=j'$$ Prove that $\left(^{(\sum_{j\in J}X_j)}\!/\!_{\mathcal R}, ^{(\sum_{j\in J}T_j)}\!/\!_{\mathcal R} \right)\approx \left(\sum_{j\in J}(^{X_j}\!/\!_{\mathcal R_j}), \sum_{j\in J}(^{T_j}\!/\!_{\mathcal R_j})\right)$
My attempt
I denote $\forall j\in J$:
$\ \begin{align*}p_j:X_j&\to^{X_j}\!/\!_{\mathcal R_j}\\x&\mapsto[x]_j\end{align*}$
and
$\begin{align*}p:\sum_{j\in J}X_j&\to(\sum_{j\in J}X_j)\!/\!_{\mathcal R}\\(x,j)&\mapsto[(x,j)]\end{align*}$
Now, I define $\begin{align*}f:\left(^{(\sum_{j\in J}X_j)}\!/\!_{\mathcal R}, ^{(\sum_{j\in J}T_j)}\!/\!_{\mathcal R} \right) & \to \left(\sum_{j\in J}(^{X_j}\!/\!_{\mathcal R_j}), \sum_{j\in J}(^{T_j}\!/\!_{\mathcal R_j})\right)\\ [(x,j)]&\mapsto([x]_j,j)\end{align*}$
I can prove that:
- $f$ is bijective: using the definition of $\mathcal R$
- $f$ is continuous: using the Fundamental Property of Product Topology, $f\circ p= (p_j \times 1_J)$.
My problem is how to prove $f$ is open. I wolud like to prove it using the Fundamental Property of Sum Topology. I can see that $\forall k\in J\ f^{-1}\circ p_k=p\circ j_k\circ p_k^{-1}$. But I can't tell that $p_k$ is open.