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We say that $X$ is a standard Borel space iff it is a Polish space equipped with the Borel $\sigma$-algebra. Similarly, a standard Borel group is a Polish group s.t. multiplication and inversion are both Borel maps. These concepts are very common in Classical Descriptive Set Theory and in order to justify these definitions, Kechris (pp. $92$, row $-3$) provides the following example:

Let $X$ be a standard Borel space and $G$ a standard Borel group acting on it as a Borel map. If $E_G$ denotes the equivalence relation induced by this action $$x E_G y \iff \exists g\in G(g.x=y),$$ it is easy to verify that $E$ is analytic in $X^2$.

How should a proof look like?

Recall that a subset $A$ of a standard Borel space $X$ is analytic iff there exists $Y$ Polish space and $f$ Borel bijection from $X$ to $Y$ s.t. $f(A)$ is analytic in $Y$.

I apologize for this low-level question, but looking at this I don't see how to put together these informations to get a proof. Thank you in advance for your help.

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The definition of an analytic subset $A$ of a standard Borel space $X$ is probably better understood in view of Kechris's Exercise 14.6: there is a standard Borel space $Z$ and a Borel function $f : Z \to X$ such that $A = f(Z)$.

Here, note that $X^2$ and $G \times X^2$ are standard Borel spaces. Consider the map $f : G \times X^2 \to X^2$ defined by $f(g,x,y) = (g \cdot x, y)$, which is a Borel function because the group action is Borel, and let $D = \{(x,x) : x \in X\} \subset X^2$ be the diagonal of $X^2$. Note that $D$ is a Borel set in $X^2$ and so $B = f^{-1}(D)$ is a Borel set in $G \times X^2$.

Now let $\pi : G \times X^2 \to X^2$ be the projection $\pi(g,x,y) = (x,y)$. This also is a Borel function, and it should be straightforward to verify that $\pi(B) = E$. Since a Borel subset of a standard Borel space is itself a standard Borel space, this shows that $E$ is analytic.

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