The resulting topology is $\tau$, if
If the action of $G$ on $X$ is continuous (i.e. the multiplication map $X\times G\to X$ is continuous):continuous) then the resulting topology is $\tau$:
Let $\tilde X$ denote $(X\times G)/\sim$, and let $\phi:X\to \tilde X:x\mapsto[x,e]$ be the identification you mentioned (with the factors $X,G$ reversed for convenience). Then $\phi$ is clearly continuous. Let $\psi:\tilde X\to X$ be the inverse of $\phi$, i.e. $\psi([x,g]) = xg$. For an open subset $U$ of $X$, $\psi^{-1}(U) = {(x,g):xg\in U, g\in G}$U}$. Pulling this back to$X\times G$via the projection$X\times G\to\tilde X$gives exactly the preimage of$U$under the multiplication map$X\times G\to G$, which is open, and so$\psi^{-1}(U)$is open in$\tilde X$. So$\phi$is a homeomorphism. 2 added 6 characters in body The resulting topology is$\tau$, if the action of$G$on$X$is continuous (i.e. the multiplication map$X\times G\to X$is continuous): Let$\tilde X$denote$(X\times G)/\sim$, and let$\phi:X\to \tilde X:x\mapsto[x,e]$be the identification you mentioned (with the factors$X,G$reversed for convenience). Then$\phi$is clearly continuous. Let$\psi:\tilde X\to X$be the inverse of$\phi$, i.e.$\psi([x,g]) = xg$. For an open subset$U$of$X$,$\psi^{-1}(U) = {(x,g):xg\in U, g\in G}$. Pulling this back to$X\times G$via the projection$X\times G\to\tilde X$gives exactly the preimage of$U$under the multiplication map$X\times G\to G$, which is open, and so$\psi^{-1}(U)$is open in$\tilde X$. So$\phi$is a homeomorphism. 1 The resulting topology is$\tau$, if the action of$G$on$X$is continuous (i.e. the multiplication map$X\times G\to X$is continuous): Let$\tilde X$denote$(X\times G)/\sim$, and let$\phi:X\to \tilde X:x\mapsto[x,e]$be the identification you mentioned (with the factors reversed for convenience). Then$\phi$is clearly continuous. Let$\psi:\tilde X\to X$be the inverse of$\phi$, i.e.$\psi([x,g]) = xg$. For an open subset$U$of$X$,$\psi^{-1}(U) = {(x,g):xg\in U, g\in G}$. Pulling this back to$X\times G$via the projection$X\times G\to\tilde X$gives exactly the preimage of$U$under the multiplication map$X\times G\to G$, which is open, and so$\psi^{-1}(U)$is open in$\tilde X$. So$\phi\$ is a homeomorphism.