Equaliser of homomorphism of integers

I have the following situation:

$\mathbb{Z} \mathop{\rightrightarrows^{0}_{2} \mathbb{Z}} \ \$

Where here $0$ and $2$ stands for the multiplication.

Now, I want to find the equaliser of this diagram (in the category of Abelian Groups), and I know it should be $\mathbb{Z}$, but I don't understand why. The matter here is that is involved the $0$-map which makes everything "trivial", so why not $\{ 0 \}$ as an equaliser?

The equalizer of $0$ and $2$ is the set of $x\in Z$ such that $0(x)=2(x)$ i.e $0=2x$, i.e $x=0$. So the equalizer is $\{0\}$.

Building on the answer above, the equalizer is literally that: the two paths are equal.

So ask yourself: when is 0 times a number the same as 2 times that number? As the answer above shows, only when 0 is that number.

Now extend that.

Imagine if the paths were 1 and 2 instead of 0 and 1. What is the equalizer then? Well, we need a number in Z such that 1 times that number equals 2 times that number. Is there such a number in Z?

Now imagine that the paths were 2 and 3. Now we are looking for a number in Z such that 2 times that number equals 3 times that number? Is there such a number in Z?

Thus any two paths in Z have no equalizer unless {0}, Z's kernal, is one of the paths.