Induced differential operator in long exact sequence of cohomology groups of differential complexes is well defined

Question

In Differential Forms in Algebraic Topology, Bott & Tu they claim that given a short exact sequence of differential complexes:

$$0\rightarrow A \overset{f} \rightarrow B \overset{g} \rightarrow C \rightarrow 0$$ with $f,g$ being chain maps, there is a long exact sequence of cohomology groups:

They define the induced differential operator $d^{\ast}$ as follows. Given $c\in\ker d \cap C^{q}$ we have a $b\in B\cap B^q$ such that $g(b)=c$ and $db\in \ker g = \text{Im} f$. So, we have an $a\in A^{q+1}$ such that $f(a)=db$. We define $d^{\ast}([c])=[a]\in H^{q+1}(A)$.

They say that a simple diagram chasing shows that $d^{\ast}$ is well-defined on equivalence classes by for some reason I'm not able to see it. I've tried to prove that if $c'-c \in \text{Im} d \cap C^{q}$ such that $g(b)=c, \ g(b')=c'$ and $f(a)=db, \ f(a')=db'$ then $a'-a\in \text{Im} d \cap A^{q+1}$ but I am having a lot of trouble showing this. I would appreciate some hints about what I'm missing here.

Answer 1

The given short exact sequence of differential complexes - and i assume we have complexes of modules over a fixed ring, so that diagram chasing is possible - looks in detail as follows:

$\require{AMScd}$ $$ \begin{CD} @. \vdots @. \vdots @. \vdots \\ @. @A{d_A}AA @A{d_B}AA @A{d_C}AA \\ 0 @>>> \bbox[limegreen]{\ A^{q+1}\ } @>f>> B^{q+1} @>g>> C^{q+1} @>>> 0\\ @. @A{d_A}AA @A{d_B}AA @A{d_C}AA \\ 0 @>>> A^{q} @>f>> B^q @>g>> \bbox[yellow]{\ C^q\ } @>>> 0\\ @. @A{d_A}AA @A{d_B}AA @A{d_C}AA \\ 0 @>>> A^{q-1} @>f>> B^{q-1} @>g>> C^{q-1} @>>> 0\\ @. @A{d_A}AA @A{d_B}AA @A{d_C}AA \\ @. \vdots @. \vdots @. \vdots \\ \end{CD} $$

We define as in the question the map from the homology computed in the yellow entry, with values in the homology computed in the green entry:

• Pick some $\bbox[yellow]{\ c\ }$ in the yellow object.
• Since $g=g^q$ is surjective, lift $c$ to some $b\in B^q$.
• Build $db=d_Bb$, an element in $B^{q+1}$. This element is mapped to $0$ via $g$, so by exactness, it comes via $f$ from an element (which is unique) $\bbox[limegreen]{\ a\ }\in A^{q+1}$, determined via $fa=db$
• We define the connecting morphism, usually denoted by $\delta:H^\cdot(C)\to H^{\cdot+1}(A)$, as a map $c\to a$, more exactly, the class of $c$ goes to the class of $a$.

Why is it well defined?

• Why the $g$-lift $b$ of $a$ leads to no ambiguity? Pick an other lift, $b'$ of $c$ in $B^q$. We built the image $db'$ of $b'$ in $B^{q+1}$. Then pick $a'\in A^{q+1}$ so that $fa'=db'$. It remains to check that $a$ and $a'$ coincide as elements in $H(a)$. Indeed, from $g(b-b')=gb-gb'=c-c=0$, and from the exactness of the horizontal short exact sequence in level $q$ we obtain an $\alpha in A^q$ with $$ f\alpha = b-b'\ . $$ Then: $$ a-a'=d\alpha\ . $$ Why? The above relation is equivalent to the relation obtained from it by applying the injective / monomorphic $f$, and indeed: $$ f(a-a')=fa-fa'=db-db'=d(b-b')=d\; f\alpha =f\; d\alpha\ . $$ So $\delta$ is well defined w.r.t. the choice of a $g$-lift.
• Why are boundaries mapped to zero? Assume we start with some $c=d\kappa$, $\kappa\in C^{q-1}$. Then we can lift $\kappa$ via $g$ in level $q-1$ to some $\beta$, so $g\beta=\kappa$. This means that $d\beta$ is a possible lift for $c=d\kappa$. We take it, any lift is equally good, we have cleared this above. We move to $B^{q+1}$ via $d$ and get $dd\kappa=0$, which uniquely lifts to $0\in A^{q+1}$. So boundaries in $C^q$ are mapped to zero.

$\square$