Assume we have a the $\mathbb Z\text{ modules}$ $S1 \equiv (F, E, V)$ with boundary maps $(\partial_{FE}: F \rightarrow E$, $\partial_{EV}: E \rightarrow V)$, with the condition that $\partial_{EV} \circ \partial_{FE} = 0$ (homology condition). We interpret this as the description of a simplicial complex.
Similarly, we have another space $S_2 \equiv (F', E', V')$ with associated boundary maps $\partial_{F'E'}: F' \rightarrow E', \partial_{E'V'}: E' \rightarrow V'$ which also satisfy the homology condition $\partial_{E'V'} \circ \partial_{F'E'} = 0$.
Now, I describe a "mapping" between these two simplicial complexes, given by $(f: F \rightarrow F', e: E \rightarrow E', v: V \rightarrow V')$, such that the diagram commutes:
$$ \begin{bmatrix} F & \xrightarrow{\partial_{FE}} & E &\xrightarrow{\partial_{EV}} & V \\ f\downarrow & & e\downarrow & & v\downarrow \\ F' & \xrightarrow{\partial_{F'E'}} & E' &\xrightarrow{\partial_{E'V'}} & V' \\ \end{bmatrix} $$
At this point, I would like to conclude that the mapping $(f, e, v)$ can only simplify the homology group. Formally, I wish to show that:
$$ \text{there exists a surjective abelian group homomorphism } \phi: Ker(\partial_{EV})/Im(\partial_{FE}) \twoheadrightarrow Ker(\partial_{E'V'})/Im(\partial_{F'E'}) $$
I feel like I know nothing in the theory of exact sequences that will let me quickly prove this. However, I feel strongly that there must be a gadget that lets me get this quickly? I'm terribly sorry if this is something 'standard'. I self taught myself this material, so my understanding is quite scattered.