Obtaining the $0$th cohomology group from the Universal Coefficient Theorem

Question

Suppose I have a sequence of abelian groups $\cdots C_2\to C_1\to C_0\to 0$. Then the Universal Coefficient Theorem tells us that for any abelian group, $A$, we have the exact sequence $$0\to Ext^1(H_{j-1}(C),G)\to H^j(C;G)\to Hom(H_j(C),G)\to 0$$ So, if we know the homology groups, then we can find $H^j(C;G)$ for $j\geq 1$. However, can we use this to obtain $H^0(C;G)$ as well? Would, for example, it be valid to assert $Ext^1(H_{-1}(C),G):=0$?

Answer 1

This is true since all negative singular cohomology is trivial.