$6 \cdot 140$ Find all functions $f: Q \rightarrow$ $Q$ (where $Q$ is the set of rational numbers) that satisfy $f(1)=2$ and
$f(x y) \equiv f(x) f(y)-f(x+y)+1, x, y \in Q$.
[Solution] In the original identity, let $y=1$, then we get the identity
$$
f(x) \equiv f(x) f(1)-f(x+1)+1, x \in \mathbb{Q}.
$$

That is,
$$
f(x+1) \equiv f(x)+1.
$$

Thus, for all $x \in \mathbb{Q}, n \in \mathbb{Z}$, we have $f(x+n)=f(x)+n$.
Therefore,
$$
f(n)=f(1)+n-1=n+1.
$$

Next, in the original identity, take $x=\frac{1}{n}, y=n, x \in \mathbb{Z}$, we get
$$
f\left(\frac{1}{n} \cdot n\right)=f\left(\frac{1}{n}\right) \cdot f(n)-f\left(\frac{1}{n}+n\right)+1,
$$

Thus $2=f\left(\frac{1}{n}\right)(n+1)-f\left(\frac{1}{n}\right)-n+1$,
which means $f\left(\frac{1}{n}\right)=1+\frac{1}{n}$.
Finally, take $x=p, y=\frac{1}{q}, p \in \mathbb{Z}, q \in \mathbb{N}$, we get
$$
f\left(p \cdot \frac{1}{q}\right)=f(p) \cdot f\left(\frac{1}{q}\right)-f\left(p+\frac{1}{q}\right)+1.
$$

This gives $f\left(\frac{p}{q}\right)=(p+1)\left(\frac{1}{q}+1\right)-\frac{1}{q}-p$
$$
=\frac{p}{q}+1.
$$

Therefore, the only function $f(x)=x+1$ satisfies all the conditions in the problem.