How to embed $(X,\mathcal T)$ in the product topology $\prod_{i\in I}(X,\mathcal T_i) $ For each $i\in I$, $\mathcal T_i$ is a topology on $X$ and $\mathcal T$ is the coarsest topology containing all $\mathcal T_i$.
Can we embed $(X,\mathcal T)$ (homeomorphically) in the product topology $\prod_{i\in I}(X,\mathcal T_i) $?
(we assume $I\ne \emptyset$)
 A: You have an obvious continuous injective map $(X,\mathcal T)\to\prod_{i\in I}(X,\mathcal T_i)$, namely the diagonal map. If $X$ is compact, then this map should be a homeomorphism onto its image, but not in general.
A: The diagonal map is a homeomorphism. For each non-empty finite $F\subseteq I$ let $$\mathscr{B}_F=\left\{\bigcap_{i\in F}U_i:U_i\in\mathscr{T}_i\text{ for each }i\in F\right\}\;,$$
and let $\mathscr{B}=\bigcup\{\mathscr{B}_F:F\subseteq I\text{ is finite and non-empty}\}$; then $\mathscr{B}$ is a base for $\mathscr{T}$.
For $i\in I$ let $X_i$ denote the space $\langle X,\mathscr{T}_i\rangle$, and let $d:X\to\prod_{i\in I}X_i:x\mapsto\hat x$ be the diagonal map: $\hat x_i=x$ for each $i\in I$. Let $\Delta=d[X]$ be the diagonal in $\prod_{i\in I}X_i$. For $x\in X$ let $\hat x=d(x)$.
Suppose that $F\subseteq I$ is finite and non-empty, and $B=\bigcap_{i\in F}U_i\in\mathscr{B}_F$. Then for each $x\in X$ we have 
$$\begin{align*}
d[B]&=\{\hat x\in\Delta:x\in B\}\\
&=\{\hat x\in\Delta:\forall i\in F(x\in U_i)\}\\
&=\{\hat x\in\Delta:\forall i\in F(\hat x_i\in U_i)\}\\
&=\Delta\cap\left(\prod_{i\in F}U_i\times\prod_{i\in I\setminus F}X_i\right)\;,
\end{align*}$$
an open set in the topology that $\Delta$ inherits from $\prod_{i\in I}X_i$. Thus, $d$ is an open bijection from $X$ to $\Delta$. Moreover, the sets of the form
$$\prod_{i\in F}U_i\times\prod_{i\in I\setminus F}X_i\;,$$
where $F$ ranges over finite, non-empty subsets of $I$, and $U_i\in\mathscr{T}_i$ for each $i\in F$, form a base for the product topology, so the the same calculation shows that $d$ is continuous and therefore a homeomorphism.
