Vtyjkyuk

Let $mathscr C$ be a Abelian category, how to show a morphism $varphi^•: A^•to B^•$ of complexes induce the morphism of cohomology objects $H^i(A^•)to H^i(B^•)$?

$A^•:qquad cdotsto A^i-1stackrelf^i-1toA^istackrelf^itoA^i+1stackrelf^i+1tocdots$

$B^•:qquad cdotsto B^i-1stackrelg^i-1toB^istackrelg^itoB^i+1stackrelg^i+1tocdots$

However, I can only give the morphism $operatornamekerf^ito operatornamekerg^i$.

Let $varphi^bullet : (A^bullet, d_A^bullet)to (B^bullet,d_B^bullet)$ be a morphism of cochain complexes. Concretely, recall that this means that for any $i,$ we have $d_B^icircvarphi^i = varphi^i+1circ d_A^i$ as maps $A^ito B^i+1.$

1. First, observe that $varphi^i$ induces a morphism $varphi^i : ker d_A^itoker d_B^i$ by restriction. Given $alphainker d_A^isubseteq A^i,$ we have $d_B^icircvarphi^i(alpha) = varphi^i+1circ d_A^i(alpha) = varphi^i+1(0) = 0.$ This means that $varphi^i(alpha)inker d_B^i,$ which is what we wanted.


2. In order to show that we obtain a well-defined map at the level of cohomology, we must show that the composition
   $$
   ker d_A^ixrightarrowvarphi^iker d_B^ixrightarrowpiker d_B^i/operatornameimd_B^i-1 = mathrmH^i(B^bullet)
   $$
   factors through $H^i(A^bullet);$ i.e., that any $alphainoperatornameimd_A^i-1$ is sent to $0$ in the quotient. Precisely, this means that we must check that such an $alpha$ is mapped into $operatornameimd_B^i-1$ under $varphi^i.$ To that end, let $alphainoperatornameimd_A^i-1,$ and write $alpha = d_A^i-1(beta)$ for some $betain A^i-1.$ Then it follows that
   $$
   varphi^i(alpha) = varphi^i(d_A^i-1(beta)) = d_B^i-1(varphi^i-1(beta))inoperatornameimd_B^i-1.
   $$
   This shows that the map $picircvarphi^i : ker d_A^ito mathrmH^i(B^bullet)$ factors through a unique map $(varphi^i)^ast : mathrmH^i(A^bullet)tomathrmH^i(B^bullet).$ This is the desired map.