English
For f and g, and φ, ψ, there is a natural equality mapShortComplexH1(g ∘ f, φ ≫ Res(f).map ψ) = mapShortComplexH1(f, φ) ≫ mapShortComplexH1(g, ψ).
Русский
Существует естественное равенство отображений между короткими комплексами: mapShortComplexH1(g ∘ f, φ ≫ Res(f).map ψ) = mapShortComplexH1(f, φ) ≫ mapShortComplexH1(g, ψ).
LaTeX
$$$\mathrm{mapShortComplexH1}(g \cdot f, \phi \gg (\mathrm{Res}\_f(\psi))) = (\mathrm{mapShortComplexH1} f \phi) \cdot (\mathrm{mapShortComplexH1} g \psi)$$$
Lean4
@[reassoc (attr := simp), elementwise (attr := simp)]
theorem map_id_comp_H0Iso_hom {A B : Rep k G} (f : A ⟶ B) :
map (MonoidHom.id G) f 0 ≫ (H0Iso B).hom = (H0Iso A).hom ≫ (coinvariantsFunctor k G).map f :=
by
rw [← cancel_epi (H0π A)]
ext
simp