If $\{(X_i,\mathcal{T}_i):i\in I\}=\{(X'_j,\mathcal{T}'_j):j\in J\}$ the $\Pi_{i\in I} X_i$ is homeomorphic to $\Pi_{j\in J}X'_j$ 
Statement

Let be $I$ and $J$ two collections of indices such that $|I|=|J|$. So if $\mathfrak{X}=\{(X_i,\mathcal{T}_i):i\in I\}$ and $\mathfrak{X}'=\{(X'_j,\mathcal{T}'_j):j\in J\}$ are two collections of topological spaces such that $\mathfrak{X}=\mathfrak{X}'$, then $X:=\Pi_{i\in I}X_i$ is homeomorphic to $X':=\Pi_{j\in J}X'_J$.


Proof.  Since $\mathfrak{X}=\mathfrak{X}'$ then for any $i\in I$ there exist $j_i\in J$ such that $X_i=X_{j_i}$ and so, since $|I|=|J|$, the function $\phi:I\rightarrow J$ defined by the condiction
$$
\phi(i):=j_i
$$
is a bijection; so the function $\varphi: X'\rightarrow X$ defined by the condiction
$$
[\varphi(x')](i):=x'(j_i)
$$
show that if $x'\in X'$ then $x'\in X$ and so $X'\subseteq X$; but conversely if $X'=X$ then for any $j\in J$ ther exist $i\in I$ such that $X_j=X_{i_j}$ and so, since $|J|=|I|$, then the function $\psi:J\rightarrow I$ defined by the condition
$$
\psi(j):=i_j
$$
is a bijection; so the function $\sigma:X\rightarrow X'$ defined by the condition
$$
[\sigma(x)](j):=x(i_j)
$$
show that if $x\in X$ then $x\in X'$ and so $X\subseteq X'$ and so $X=X'$ and so $X$ and $X'$ are homeomorphic through the identity.
So is the proof correct? If not how to prove the statement? Then I doubt that the function $\phi$ and $\psi$ are not bijections and that the functions $\varphi$ and $\sigma$ do not show that $X'\subseteq X$ and $X\subseteq X'$: so if these things are true I ask to prove it.
So could someone help me, please?
 A: This is false as stated, but you probably mean something else.
E.g. let $I=\Bbb N^+_0$ and $J=\Bbb N^+$ and let $\mathfrak{X}$ be indexed as follows: $X_0 = \Bbb R$, $X_1  = \Bbb R$, usual topology and all other $X_i = \{0,1\}$, discrete topology.
$\mathfrak{X'}$ is indexed by $J$ as follows: $X'_1=\Bbb R$ usual topology, $X'_i=\{0,1\}$ discrete topology.
Then $|I|=|J|$ and $\mathfrak{X}=\mathfrak{X'}$ (both are two element sets, a two point discrete one and the reals; the duplicates we don't see any more) but the first product is just $\mathbb R^2 \times C \not\simeq \Bbb R \times C$, where $C$ is the Cantor set.
What you do mean is that there is a bijection $\phi: I \to J$ such that $(X_i, \mathcal{T}_i) = (X'_{\phi(i)}, \mathcal{T}'_{\phi(i)})$.
In that case we can define a map $\Psi: \prod_{i \in I} X_i \to \prod_{j \in J} X'_j$ by $$(\Psi(f))(j) = f(\phi^{-1}(j))$$ (seeing elements of the products as functions on $I$ resp. $J$). Check that this is well-defined.
This implies that $$ \forall j \in J: \pi'_j \circ \Psi = \pi_{\phi^{-1}(j)}$$ and the universal mapping theorem for products implies that $\Psi$ is continuous.
The inverse is defined by $$\Phi(f)(i)= f(\phi(i))$$ which implies similarly that
$$\forall i \in I: \pi_i \circ \Phi = \pi'_{\phi(i)}$$ and the inverse is also continuous.
A: It’s not enough to assume that $\mathfrak{X}=\mathfrak{X}'$ and $|I|=|J|$. Let $Y$ be the one-point space, and let $E_1$ denote $\Bbb R$ with the usual topology. Let $I=J=\Bbb N$, let $\langle X_0,\mathscr{T}_0\rangle$, $\langle X_1,\mathscr{T}_1\rangle$, and $\langle X_0',\mathscr{T}_0'\rangle$ all be $E_1$, and let the spaces $\langle X_n,\mathscr{T}_n\rangle$ for $n\ge 2$ and $\langle X_n',\mathscr{T}_n'\rangle$ for $n\ge 1$ all be $Y$. Then $\mathfrak{X}=\{E_1,Y\}=\mathfrak{X}'$, but $X$ is homeomorphic to $\Bbb R^2$ with the usual topology, while $X'$ is homeomorphic to $E_1$, which is not homeomorphic to $\Bbb R$ with the usual topology. You really do need the stronger assumption that $\mathfrak{X}$ and $\mathfrak{X}'$ are equal as indexed sets, not just as sets, meaning that there is a bijection $\varphi:I\to J$ such that $\langle X_{\varphi(i)}',\mathscr{T}_{\varphi(i)}'\rangle=\langle X_i,\mathscr{T}_i\rangle$ for each $i\in I$.
This is just as well, since your attempt to construct such a bijection doesn’t quite work: in my example above, for instance, you could set $j_0=j_1=0$ and $j_n=n$ for $n>1$, and the map $i\mapsto j_i$ would satisfy the condition that $\langle X_i,\mathscr{T}_i\rangle=\langle X_{j_i}',\mathscr{T}_{j_i}'\rangle$ for each $i\in I$ without being a bijection between $I$ and $J$.
You can then use $\varphi$ to construct a function $f:X\to X'$ as follows: $[f(x)](j)=x(\varphi^{-1}(j))$ for each $x\in X$ and $j\in J$. This makes sense: the fact that $\varphi$ is a bijection ensures that $\varphi^{-1}(j)$ is well-defined for each $j\in $J, and the fact that $\langle X_{\varphi^{-1}(j)},\mathscr{T}_{\varphi^{-1}(j)}\rangle=\langle X_j',\mathscr{T}_j'\rangle$ ensures that $x(\varphi^{-1}(j))$ really is in $X_j'$. It’s not hard to check that $f$ is a bijection.
For $i\in I$ and $U\in\mathscr{T}_i$ let $S(i,U)=\{x\in X:x(i)\in U\}$; the family of all such sets $S(i,U)$ is a subbase for the product topology on $X$. Let $j=\varphi(i)$; then $\varphi^{-1}(j)=i$, so
$$\begin{align*}
f[S(i,U)]&=\{f(x):x(i)\in U\}\\
&=\{f(x):x(\varphi^{-1}(j))\in U\}\\
&=\{x'\in X':x'(j)\in U\}\;.
\end{align*}$$
This is an open set in $X'$, so $f$ is an open function, and a similar argument shows that $f$ is continuous and hence a homeomorphism.
