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

Question

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?

Answer 1

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 can also see that this map is a monomorphism without using the construction of general cones.

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 $f_n=\text{id}_{C_n}$.