Group presentation and homomorphism In Dummit and Foote, it is stated that when 2 groups G and H have generators $\{s_1, s_2, ... s_m\}$ and $\{r_1, r_2, ... r_m\}$, and that any relation that holds in G also holds in H, the mapping $\phi:G\rightarrow H$ characterised by $s_i \rightarrow r_i$ is a homomorphism.
I understand this is really obvious, but I can't seem to prove that this is true. Because everytime I attempt an argument, I find myself begging the question. 
What sorts of properties of $\phi$ do I need to prove to show that the above assertion is true? What I thought initially was to show that $\phi$ is well-defined. Would an induction argument work in this case?
 A: You‘ll have to look at the definition of a presentation to prove it formally:
$\left< a_1,\dots, a_n | r_1 = \dots = r_m =1 \right>:=F/N$
Where $F$ is the free group on $\{ a_1, \dots, a_n \}$ and $N$ is the normal subgroup generated by the $gr_ig^{-1}$ for $i \in {1, …, m}$ and $g \in F$.
Now we can take a look at the universal property of factor groups: For all groups $H$ and all homomorphisms $\varphi: F \rightarrow H$, if $N \subseteq \ker(\varphi)$ then there exists a unique homomorphism $\bar{\varphi}: F/N \rightarrow H$, s.t. $\varphi = \bar{\varphi} \circ \pi$. With $\pi: F \rightarrow F/N, a \mapsto aN$ the natural surjection. (I would draw a commutative diagram, but my LaTeX isn’t good enough).
This means that we only need to show that we can construct a homomorphism $\varphi: F \rightarrow H$ characterised by $\varphi(a_i)=h_i \in H$ (this is trivially a homomorphism). And that $N \subseteq \ker(\varphi)$ i.e. $\varphi(g r_i g^{-1})=1$, so it suffices to show: $\forall i: \ \varphi(r_i)=1$.
