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Let $(X,\tau)$ be a topological space and $p:X\rightarrow X/\sim$ a natural map for $X$. Assuming the topology $\tilde{\tau}$ whose sets are the subsets $\tilde{U}$ of $X/\sim$ such that $p^{-1}(\tilde{U})$ is open in $X$, I want to show that $p$ is a quotient map. (In other words, I want to confirm that this topology $\tilde{\tau}$ is the quotient topology for the map $p$ - the topology for which $p$ is a quotient map).

My attempt:

Since for any class $[x]\in X/\sim$ there is at least one element $y$ of $X$ such that $p(y)=[x]$ (just take any element of the class), $p$ is sujective. Now, by construction of $\tilde{\tau}$, one has trivially that $p$ is continuous. Finally, I must show that, if $p^{-1}(\tilde{U})$ is open in $X$, $\tilde{U}$ is open in $X/\sim$.

I do not know how to show this last bit.

EDIT: The definition of quotient topology that I'm using is from Sutherland's book: let $(X,\tau)$ be a topological space and $\sim$ a relation on $X$. Let $p:X\rightarrow X/\sim$ denote the function that maps each element of $X$ to it's equivalence class (called the natural map).The quotient topology is the family $\tilde{\tau}$ of subsets $\tilde{U}$ of $X/\sim$ such that $p^{-1}(\tilde{U})\in \tau$.

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    Aren't you "proving" a definition? – 2017-01-18
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    No. I should explain myself, then: in Sutherland's book, he defines the quotient topology without referring to the quotient map. He then goes on to define quotient map, and says that the natural map is indeed a quotient map, but does not prove it. I was trying to do so. So, in this approach, $p$ is simply a function from $X$ into $X\sim$ (no other assumptions are made), and the quotient topology is precisely the set of subsets which are pre-images of open sets, just like I wrote above. – 2017-01-18
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    @HennoBrandsma (I forgot to identify you on my last comment, and now I can't edit it). – 2017-01-18
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    How is Sutherland's definition exactly ? How does he define the topology on $X / \sim$ without referring to the quotient map? – 2017-01-18
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    @HennoBrandsma Please see my Edit. – 2017-01-18
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    It still seems restating the definition to me. – 2017-01-19
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    @HennoBrandsma So where does the "if $p^{-1}(\tilde{U})$ is open in $X$, $\tilde{U}$ is open in $X/\sim$" property come from? That's what I asked above, basically: I really can't see it. – 2017-01-19

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