Surjective group homomorphism If we have two group presentations, can we say that we have a surjective group homomorphism $f: G \rightarrow H $ if the generators of $G$ map to all the generators of $H$ and all the relations of $G$ map to all the relations of $H$ in the appropriate way? 
Thanks 
 A: Given $G=\langle g_i\mid r_j\rangle $ and $H=\langle h_k\mid s_l\rangle$, a homomorphism $f\colon G\to H$ is given by specifying $f(g_i)\in H$ such that for each relation $r_j$ you can show $f(r_j)=1$  using the relations $s_l$. The homomorphism is additionally surjective if you can exhibit for each $k$ an element $x_k\in G$ such that $f(x_k)=h_k$.
A: I am not sure what the appropriate way to map a relation to another is (the relations do not actually exist in the group, but rather in the free group on the generators that one is quotienting by in order to obtain $G$), but for surjectivity: it is sufficient for the homomorphism to map some generating set to some other generating set, and it is also necessary for it to map every generating set to another generating set. To see this, consider the following
Lemma. Let $f:G\to H$ be a homomorphism, and $X\subseteq G$ a subset. Then the image of the span of the set $X$  under the map $f$ is the span of the image of $X$ under $f$; i.e. $f^*\langle X\rangle=\langle f^*X\rangle$.
Proof. (formal enough for my personal satisfaction anyway)
$$\begin{array}{cl} f^*\langle X\rangle & =f^*\{x_1^{e_1}\cdots x_m^{e_m}:x_j\in X\} \\ & =\{f(x_1)^{e_1}\cdots f(x_m)^{e_m}:x_j\in X\} \\ & = \{u_1^{e_1}\cdots u_m^{e_m}:u_j\in f^*X\} \\ & =\langle f^*X\rangle.\end{array}$$
And hence
Corollary. Suppose $\langle X\rangle=G$, i.e. $X$ is a generating set. Then  $f:G\to H$ is surjective iff $f^*G=H$ iff $f^*\langle X\rangle=H$ iff $\langle f^*X\rangle=H$ iff $f^*X$ is a generating set for $H$.
