Given a function $f$ and an equivalence relation $\sim$ on the domain of $f$, if two equivalent elements with respect to $\sim$ have the same image under $f$, then $\sim$ is called the equivalence kernel of $f$
In other words $\sim$ is the equivalence kernel of $f$ if and only if for any $x_1,x_2 \in D_f$: $$x_1\sim x_1⇔f\left(x_{1}\right)=f\left(x_{2}\right)$$
For example the quotient mapping of a set $X$ defined as $\pi:X \to X/\sim$ is a function such that for all $x_1,x_2 \in X$: $$x_1 \sim x_2 ⇔\left[x_{1}\right]=\left[x_{2}\right]$$ or equivalently : $$x_1 \sim x_2 ⇔\pi(x_{1})=\pi(x_{2})$$
Where $ \left[a\right]$ is the equivalence class of $a$.
Using definition of an equivalence kernel it follows $\sim$ is actually the equivalence kernel of $\pi$
As Wikipedia says the equivalence kernel of an injection is the identity relation.
By definition $f:X\to Y$ is injection if and only if: $$f(x_1)=f(x_2)⇒x_1=x_2$$
the equivalence kernel of $f$ would be the equivalence relation $=$ but we also should show that the other direction $$x_1=x_2⇒f(x_1)=f(x_2)$$ holds.
I think this direction is clear, but I'm not sure about that, because maybe there is another reason for that .
And also I don't know how exactly identity relation will help us ( maybe it's for proving the other direction I'm looking for) Can someone help me with this?
For every function, we have$$x_1=x_2\implies f(x_1)=f(x_2)$$and therefore $f$ is injective if and only if$$x_1=x_2\iff f(x_1)=f(x_2).$$So, if you define $x_1\sim x_2$ as $f(x_1)=f(x_2)$, then $f$ is injective if and only if $\sim$ is the identity relation.