/sci/ - Science & Math

In fact in Lang's Algebra it says: (E is the standard resolution of Z as in the wiki-article)
"Show that if <span class="math">H^Q(G,A)[/spoiler] denotes the q-th homology of the complex Hom(E,A), then <span class="math">H^0(G,A) = A^G[/spoiler] (this is the submodule of G-invariants). Thus the left derived functors of <span class="math">A \mapsto A^G[/spoiler] are the homology groups of the complex Hom(E,A)."

Could someone explain the "thus"-part to me?

>>1958714
what on bro?
>>1958716
anti-sage