# Why does every chain complex have a map into its cone?

In Weibel's An introduction to homological algebra he defines a cone as an explicit chain complex associated to the given one -i.e. for a chain $$C=(C_i, d)$$ he defines $$Cone(C)=\left(C_{i-1} \oplus C_i, \begin{bmatrix} -d & 0 \\ -id & d \end{bmatrix} \right)$$.

It is good and it is trivial that we have a monomorphism $$C\rightarrow Cone(C)$$ which gives us a short exact sequance $$0\rightarrow C\rightarrow Cone(C)\rightarrow C[-1]\rightarrow 0$$, however we can define cone as an universal object as follows below.

For a chain complex $$C$$ let $$F$$ be a functor from chain complexes to sets given by the formula $$F(D)= \left\{(f,s); f:C\rightarrow D, f=ds+sd,\text{f is a morphism}, \text{s is a chain homotopy}\right\}$$ then the above cone represents it because $$Hom(Cone(C),D)$$ is naturally bijective to $$F(D)$$.

Assume now that our cone is defined as an universal object. Why do we have a monomorphism from $$C$$ to $$Cone(C)$$?

I don't know almost anything about a homotopical algebra but maybe it is a proposition there?

• This is justs the structural map of the universal property (i.e. it corresponds to the identity of cone(C) ). Of course you cannot see abstractly that it is a mono. Aug 9, 2019 at 11:02

This is really just rephrasing Simon Henry's comment, but there is a natural transformation $$F\to\text{Hom}(C,-)$$ given by $$(f,s)\mapsto f$$, and so if $$\text{Cone}(C)$$ represents $$F$$ then by Yoneda's lemma this natural transformation is induced by a map $$C\to\text{Cone}(C)$$.
You need to show that for every complex $$B$$ and every nonzero map $$\alpha:B\to C$$, the composition $$B\to C\to\text{Cone}(C)$$ is nonzero, or equivalently, by Yoneda, that the natural transformation $$F\to\text{Hom}(B,-)$$ given by $$(f,s)\mapsto f\circ\alpha$$ is nonzero. Or in other words, that there is some complex $$D$$ and a map $$f:C\to D$$ that is homotopic to zero with $$f\circ\alpha$$ nonzero.
But if $$\alpha$$ is nonzero in degree $$n$$, then you can take $$D$$ to be the complex $$\dots\to0\to C_n\stackrel{\text{id}_C}{\to}C_n\to0\to\dots$$ with nonzero terms in degrees $$n-1$$ and $$n$$, with $$f_{n-1}=d_{n-1}$$ and $$f_n=\text{id}_{C_n}$$.