# Defining a Map Between Two Chain Complexes

I would like someone to check my reasoning here and, if my reasoning is correct, help me define a map to make a short exact sequence.

I am given a short exact sequence of chain complexes $$0\longrightarrow B\stackrel{f}{\longrightarrow} C\stackrel{g}{\longrightarrow} D\longrightarrow 0.$$ Let $C_f$ be the cone on $f$, so $(C_f)_n=B_{n-1}\oplus C_n$. And let $Cyl_g$ be the cylinder on $g$, so $(Cyl_g)_n=C_n\oplus C_{n-1}\oplus D_n$. Then there is a chain map $$C_f\stackrel{\alpha}{\longrightarrow} Cyl_g$$ given by $\alpha(b,c)=(c,-f(b),0)$ where $c\in C_n$ and $b\in B_{n-1}$. Since $f$ is injective, $\alpha$ is injective. It seems to me that the cokernel of $\alpha$ should be $D[-1]\oplus D$, were $D[-1]$ is the shift of $D$. So, I should get a short exact sequence $$0\longrightarrow C_f\stackrel{\alpha}{\longrightarrow} Cyl_g\stackrel{\beta}{\longrightarrow} D[-1]\oplus D\longrightarrow 0.$$ My problem is defining $\beta$. I tried the obvious $\beta(c_1,c_2,d)=(g(c_2),d)$. But, when I checked, that did not define a chain map. Assuming my reasoning is correct, what should $\beta$ be? Thanks.

EDIT: The differential on $Cyl_g$ is given by $$\partial(c_1,c_2,d)=(\partial c_1+c_2,-\partial c_2,\partial d -g(c_2)).$$ Then $$\beta(\partial(c_1,c_2,d))=(-\partial g(c_2),\partial d-g(c_2))$$ and $$\partial(\beta(c_1,c_2,d))=(\partial g(c_2),\partial d).$$

• Why do you think the definition you gave of $\beta$ doesn't work? It seems reasonable to me. – Hugh Thomas Mar 14 '15 at 3:52
• @HughThomas I have edited the question to show my calculations. – Joe Johnson 126 Mar 14 '15 at 16:41
• $\alpha$ also doesn't define a chain map. Something seems to be amiss with the given information. – Hugh Thomas Mar 16 '15 at 3:07
• @HughThomas The definition for the differential on $C_f$ is $$\partial(b,c)=(-\partial b,\partial c-f(b)).$$ Given that $g(f(b))=0$, $\alpha$ is a chain map by my calculations. – Joe Johnson 126 Mar 16 '15 at 15:08

I think your choice of $\beta$ works.
However, the differential on $D_{n-1}\oplus D_n$ is given by $\partial(d_1,d_2)=(-\partial d_1, \partial d_2 -d_1)$.
• Just so I know I have the right idea: The differential is what you say because we are viewing $D_{n-1}\oplus D_n$ as a quotient of the cylinder. So, the differential is the one that is induced from the differential on the cylinder? – Joe Johnson 126 Mar 16 '15 at 23:04
• Yes. I mean, you essentially had to choose $\beta$ as you did, and then that forces the differential on you. – Hugh Thomas Mar 16 '15 at 23:46