Topology on $X$ using topology on $ X × Y $ Suppose there is topology on set $X × Y$. This is not product topology using topology on $X$ and topology on $Y$, all we know is $X$ and $Y$ are just sets. Then can we put topology on $X$? 
 A: Let $\tau$ be the topology on $X\times Y$, and let $\pi:X\times Y\to X:\langle x,y\rangle\mapsto x$ be the projection to the first factor. Let $\tau_X=\{U\subseteq X:\pi^{-1}[U]\in\tau\}$. If $U,V\in\tau_X$, then
$$\pi^{-1}[U\cap V]=\pi^{-1}[U]\cap\pi^{-1}[V]\in\tau\;,$$
so $U\cap V\in\tau_X$. Similarly, if $\mathscr{U}\subseteq\tau_X$, then
$$\pi^{-1}\left[\bigcup\mathscr{U}\right]=\bigcup\{\pi^{-1}[U]:U\in\mathscr{U}\}\in\tau\;,$$
so $\bigcup\mathscr{U}\in\tau_X$. Clearly $\varnothing,X\in\tau_X$, so $\tau_X$ is a topology on $X$ determined by the given topology $\tau$ on $X\times Y$.
Note that $\tau_X$ may not be very nice. For instance, for $\alpha\in\Bbb R$ let 
$$B_\alpha=\{\langle x,y\rangle\in\Bbb R^2:x+y=\alpha\}\;,$$
and let $\mathscr{B}=\{B_\alpha:\alpha\in\Bbb R\}$; $\mathscr{B}$ is a base for a topology $\tau$ on $\Bbb R^2$, and it’s not hard to check that in this case $\tau_{\Bbb R}=\{\varnothing,\Bbb R\}$, the indiscrete topology on $\Bbb R$.

As an exercise you might find it instructive to try to prove that $\left\langle\Bbb R^2,\tau\right\rangle$ is homeomorphic to $X\times Y$, where $X$ is $\Bbb R$ with the discrete topology, and $Y$ is $\Bbb R$ with the indiscrete topology.

A: There is a natural surjective map from $X \times Y$ to $X$, namely the projection function $p(x,y)=x$. 
So, it seems natural to take the quotient topology on $X$ induced by the given topology on $X \times Y$ and the function $p$. 
Briefly, this quotient topology is defined to be the strongest topology on $X$ such that $p$ is a continuous function (with respect to the given topology on $X \times Y)$. In more detail, this quotient topology on $X$ is defined so that a subset $U \subset X$ is open if and only if $p^{-1}(U) = U \times Y \subset X \times Y$ is open.
By the way, when using given topologies on $X$ and $Y$ and the product topology on $X \times Y$, this is the way that you can recover the given topology on $X$: it is the quotient of the product topology on $X \times Y$ with respect to the projection function $p : X \times Y \to X$.
Also, there's nothing special about Cartesian products in this discussion. Given any topological space $Z$, any set $X$, and any surjective function $p : Z \to X$, the quotient topology on $X$ is a natural topology associated to $Z$ and $p$.
