$6 \cdot 140$ Find all functions $f: Q \rightarrow Q$ (where $Q$ is the set of rational numbers) satisfying $f(1)=2$ 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

that is $\square$
$$f(x) \equiv f(x) f(1)-f(x+1)+1, x \in Q .$$
$$f(x+1) \equiv f(x)+1$$

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

Next, in the original identity, take $x=\frac{1}{n}, y=n, x \in Z$, 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 $x=p, y=\frac{1}{q}, p \in Z, q \in N$, 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 leads to $f\left(\frac{p}{q}\right)=(p+1)\left(\frac{1}{q}+1\right)-\frac{1}{q}-p$
$$=\frac{p}{q}+1$$

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