Let $I,J$ be sets and $\mathbf{r}=(r_{j})_{j\in J}$ a family of elements of the free group $F(I)$. For $i\in I$ let $\phi_i$ denote the canonical homomorphism of $\mathbb{Z}$ into $F(I)$.
Let $$N(\mathbf{r}):=\bigcup_{n\in\mathbb{N}}\left\{y\ |\ (\exists g)(\exists a)(g\in F(I)\land a\in\left(\bigcup\left\{r_j,r_j^{-1}\right\}\right)^{[1,n]}\land y=\prod_{i=1}^nga_ig^{-1})\right\};$$ in other words, $N(\mathbf{r})$ is the normal subgroup of $F(I)$ generated by the $r_j$. Let $\pi:F(I)\rightarrow F(I)/N(\mathbf{r})$ be the canonical homomorphism. For $i\in I$ let $\tau_i=\pi(\phi_i(1))$. Then $(\tau_i)_{i\in I}$ is a family of elements of $F(I)/N(\mathbf{r})$. Then the group $F(I)/N(\mathbf{r})$ is defined by the generators $\tau_i$ and the relators $r_j$.
Now, suppose $I=[1,n]$ and $J=[1,m]$. Let $r_j=u_j^{-1}v_j$ for $u_j,v_j\in F(I)$ and $j\in J$. Then we say that $F(I)/N(\mathbf{r})$ is defined by the presentation $$\langle\tau_1,\ldots,\tau_n;u_1=v_1,\ldots,u_m=v_m\rangle.$$
I don't understand why we write $u_j=v_j$ ($j\in J$) in the symbol above? Isn't it rather the case that the equivalence classes of $u_j,v_j$ are equal: i.e. $\pi(u_j)=\pi(v_j)$?