# More Ab. Cat. stuff and mapping cone

This post covers the basics of cochain complexes and shows roughly that the category of chain complexes with homotopy classes of morphisms form a triangulated category (even though I don’t define triangulated category).

Setup: $\mathscr{A}$ is an abelian category.  $\mathscr{C(A)}$ is the category of complexes with terms in $\mathscr{A}$; this is also an abel. cat.  The convention is a complex has increasing indices $... \to A^{q-1} \to A^q \to A^{q+1} \to ...$.  It has cohomology functors $H^q \colon \mathscr{C(A)} \to \mathscr{A}$

$H^q(A^\bullet, d^\bullet) = \ker d^q /\mbox{im } d^{q-1}$

And maps between complexes are just collection of maps between the groups in each degree (that in particular make commuting squares).

$\begin{array}{ccc} A^q & \xrightarrow{d^q_A} & A^{q+1} \\ \downarrow & \mbox{} & \downarrow \\ B^q & \xrightarrow{d_B^q} & B^{q+1} \end{array}$

There is a shift operator on complexes

$(A[m])^q = A^{m+q}$

$d_{A[m]} = (-1)^m d_A$

The $(-1)^m$ factor is introduced to make the mapping cone (see below) fit into a long sequence of complexes that gives a long exact sequence of cohomology.  And for two maps $f,g \colon A^\bullet \to B^\bullet$ its said $f,g$ are homotopic if there are maps $r_q \colon A^q \to B^{q-1}$ such that

$f^q - g^q = r^{q+1}\circ d_A^q + d_B^{q-1} \circ r^q$

This implies in particular that $f,g$ induce the same map on cohomology.  Also when $f$ induces an isomorphism on cohomology then $f$ is a quasi isomorphism.  In the case $\mathscr{A}$ has infinite products, here is another construction.  For two complexes $A^\bullet, B^\bullet$ define another complex $\hom_{A,B}$ by

$\hom_{A,B}^q = \prod_i \hom(A^i, B^{i+q})$

With coboundry map $d^q \colon \hom_{A,B}^q \to \hom_{A,B}^{q+1}$ given by

$d^q(h) = d_B \circ h + (-1)^{q+1}h \circ d_A$

A check that $d^2 = 0$:

$d^{q+1}(d^q(h)) = d_B \circ (d^q(h)) + (-1)^{q+2}d^q(h) \circ d_A$

$= d_B \circ (d_B \circ h + (-1)^{q+1}h \circ d_A) + (-1)^{q+2} d^q(h) \circ d_A$

$(-1)^{q+1} d_B \circ h \circ d_A + (-1)^{q+2}d^q(h) \circ d_A$

$(-1)^{q+1} d_B \circ h \circ d_A+(-1)^{q+2} d_B \circ h \circ d_A = 0$

Now generally the important thing about complexes are their cohomology groups, so its natural to consider the the homotopy category $\mathscr{K(A)}$ which has the same objects, but the morphisms are homotopy classes of maps.  This is not an abelian category, but it is triangulated.

### The mapping cone in K(A)

I’m not going to go through the definition of triangulated category.  A good place where this is done is the first chapter of Huybrechts.  Instead, here are some properties of $\mathscr{K(A)}$.  Say $u\colon A^\bullet \to B^\bullet$ is a morphism of complexes.  Then the mapping cone can be constructed as

$C^q = \mbox{Cone}(u) := B^q \oplus A^{q+1}$

$d_C(b,a) = (d_B(b) - u(a), -d_A(a) )$

Ogus noted in the literature, often $d_B(b) - u(a)$ is replaced by $d_B(b) + u(a)$.  In either case, it can be checked that $d^2_C = 0$.  There is natural inclusion $B^\bullet \xrightarrow{i} C^\bullet$ and natural projection $C^\bullet \to (A[1])^\bullet$.  There are indeed maps of complexes:

$d_C \circ i(b) = d_C(b, 0) = (d_B(b), 0 ) = i \circ d_B(b)$

$d_{A[1]} \circ p (b,a) = d_{A[1]}(a) = -d_A(a) = p(d_B(b) - u(a),-d_A(a)) = p \circ d_C (b,a)$

So there is a sequence

$A^\bullet \xrightarrow{u} B^\bullet \xrightarrow{i} C^\bullet \xrightarrow{p} A^\bullet [1] \xrightarrow{u[1]} B^\bullet [1]$

I can take cohomology of this sequence, and noting that $H^q(A^\bullet [1]) = H^{q+1} A^\bullet$ explains

Prop: There is an exact sequence

$H^q(A^\bullet ) \to H^q(B^\bullet) \to H^q(C^\bullet) \to H^{q+1}(A^\bullet) \to H^{q+1}(B^\bullet)$

proof: Its straightforward to check that the composition of two consecutive maps is zero, so the image is contained in the kernel.  It remains to show the kernel is contained in the image.  To show this at $H^q(B^\bullet)$ let $b$ be in the kernel, this means

1. $d_B(b) = 0$
2. $i(b) = (b, 0 ) = d_C(b',a') = (d_B(b') - u(a'), -d_A(a')$

In particular, I conclude $d_A(a') = 0$ (i.e. this is cocycle) and $b$ differs from $u(-a)$ by a coboundry, so $b$ comes from something in the image.  Now suppose $(b,a)$ is in the kernel of the map out of $H^q(C^\bullet)$.  This means

1.  $d_C(b,a) = (d_B(b) - u(a), -d_A(a) ) = (0, 0)$
2. $i(b,a) = a = d_A(a')$

To show $(b,a)$ is in the image of the previous map, I must show it differs from an element of the form $(b', 0)$ by a coboundry and that  $d_B(b') = 0$.  Now

$(b,a) + d_C(0,a') = (b - u(a'), a -d_A(a')) = (b - u(a'), 0)$

and $d_B(b - u(a')) = d_B(b) - du(a') = d_B(b) - u(da') = d_B(b) - u(a) = 0$ as required.

QED

Here is another result concerning the mapping cone

Prop: Say $0 \to A^\bullet \xrightarrow{u} B^bullet \xrightarrow{m} Q^\bullet \to 0$ is a s.e.s then defining $\pi \colon C^\bullet \to Q^\bullet$ by $\pi(b,a) = m(b)$ then $\pi \circ d_C = d_Q \circ \pi$ and $\pi$ is a quasi isomorphism.

proof: The first statement is just checking the definition of the maps and wont be done here.  For the second statement, its simply a matter of checking injectivity and surjectivity.  So suppose $(b,a) \in H^q (C^\bullet)$ maps to zero in $H^q(Q^\bullet)$.  This means

1.  $d_C(b,a) = (d_B(b) - u(a), -d_A(a)) = 0$
2. $\pi(b,a) = m(b)$ is coboundry so $m(b) = d_Q(q') = d_Q(m(b'))$ (by surjectivity of $B^\bullet \to Q^\bullet$)

To show injectivity, I need $(b,a)$ is a coboundry, i.e.$(b,a) = d_C(b',a') =$latex (d_B(b’) – u(a’), -d_A(a’))&s=-1$&s=-1$.   Now $m(b) = d_Q(m(b')) = m(d_B(b'))$ and as $m\circ u = 0$ and exactness of the s.e.s I can write

$b = d_B(b') - u(a')$.

From 1, I have $d_B(b) = u(a)$.  From the above, I have $d_B(b) = d_B(-u(a')) = u(-d_A(a')) = u(a)$, and by injectivity

$a = -d_A(a')$

Surjectivity is easier, let $q' = p(b) \in H^q(Q^\bullet)$, this means $m(d_B(b)) = d_Q(m(b)) = 0$.  From exactness, $d_B(b) = u(a)$, then consider the element $(b,a)$.  I have $d_C(b,a) = (d_B(b) - u(A), -d_A(a))$, now $u(d_A(a)) = d_B(u(a)) = 0$, but $u$ is injective, so $d_A(a) = 0$.  It follows that $(b,a)$ is a lift of $q'$

QED

Next Ogus gave the definition of triangulated category, as I said I’m not doing that here.  But the point is $\mathscr{K(A)}$ is a triangulated category with mapping cone sequences forming distinguished triangles, and the shift operator playing its usual role.