Bilinear map + commitment Let $\mathbb{G}_1,\mathbb{G}_2,\mathbb{G}_T$ be yclic group of the same order and $ e: \mathbb{G}_1 \times \mathbb{G}_2\rightarrow \mathbb{G}_T$, such that $u\in \mathbb{G}_1, g \in \mathbb{G}_2, a,b,r \in \mathbb{Z}$. 

Question 1: is the following equation correct? Is there any situation under which it might be correct? If not why? 

$e(u^a,\frac {u^b}{u^a} g)=e(u^b,g)$



Scenario: assume we have two parties: $A$ and $B$, where $A$ has two messages $m, m'$ and $B$ has a message $m$. $A$ commitments to a message $m$ as: $(r.u^m)^a$, and B commitments the same message as $(r.u^m)^b$. They send their commitments to a server. It's clear that if $A$ gives $g^{\frac{1}{a}}$ to $B$, $B$ can compute $w=g^{\frac {b}{a}}$, and send it to the server to let the server check whether two messages $m$ for different parties are equal: $e((r.u^m)^a,g^{\frac {b}{a}})=e((r.u^m)^b,g)$

Question 2: Given $g^{\frac{1}{a}}$ can B generate $w'$  to convince the server that it has commited to $m'$ rather than $m$?

All operation are $\bmod p$ where $p$ is a large prime number.
If you define $u$ as element of $G_1$, then you can't just use it in $G_2$ (in the first equation), like you did in your equation. Besides, you missed to state, if $u$ and $g$ are generators of those groups. The following holds:

$$e(u^a,g^{\frac{b}{a}}) = e(u,g)^b = e(u^b,g)$$

Anyway, you didn't do that in the actual question, where I don't really see the connection to your construction. Btw, how do A and B know that they are using the same value $r$? And then: How can the server relate anything $B$ sends to $m'$, if the server has no knowledge of $m'$ or a commitment to $m'$? So no, he can't. Because $m'$ is not used in your construction at all.

The question now is: What is your actual goal? Should the server just be able to make sure, A and B committed to the same value? And A and B have to interact to generate the witness for that? You don't need pairings for that.

edit: Maybe even Alice could cheat in the system, by not sending $g^a$ to Bob, but something else. What happens then? Well, if the server check returns true, the "commited" messages are not equal. Too many things that can go wrong, especially with that $r$, which doesn't make sense.