# Motivation for the mapping cone complexes

I was reading some topics in Homological Algebra when I came across the concepts of cone of a map of complexes and cylinder.

My knowledge of Algebraic Topology is pretty basic so I only used these concepts in a pure algebraic setting. What is the motivation for this?

The shapes of a cone for example appears only in the simplicial context of Algebraic Topology or is it possible to "see" the cone in algebraic terms ?

• The mapping cone of a morphism of chain complexes is a homotopy-theoretic generalisation of the cokernel. – Zhen Lin Feb 22 '16 at 17:35

One possible motivation for the mapping cone is the fact that a morphism of chain complexes is a quasi-isomorphism iff its mapping cone has vanishing homology. So in this sense, the homology of the mapping cone of $$f$$ measures the default of $$f$$ to be a quasi-isomorphism.

From an abstract homotopy theory point of view, one can first consider the mapping cylinder of $$f : X_* \to Y_*$$. In algebraic topology, the mapping cylinder of $$f : X \to Y$$ is $$Y \cup_{X \times 0} X \times I$$. We'll try to see how this could be translated in homological algebra, and hopefully the picture will be clearer.

In homological algebra, the interval $$I = [0,1]$$ is replaced by the chain complex $$I_*$$ that has $$I_0 = \mathbb{Z} v_+ \oplus \mathbb{Z} v_-$$ (this is a free abelian group of rank two) and $$I_1 = \mathbb{Z} e$$, $$I_n = 0$$ if $$n \neq 0,1$$, and $$d : I_1 \to I_0$$ is given by $$d(e) = v_+ - v_-$$. It's an acyclic chain complex that represents an interval (in some sense that can be made precise; it is a path object in the model category of chain complexes). Roughly speaking $$v_+$$ is the vertex $$\{1\}$$, $$v_-$$ is the vertex $$0$$, and $$e$$ is the edge between the two.

The product $$X \times I$$ becomes the tensor product $$X_* \otimes I_*$$, which has: $$(X \otimes I)_n = X_n \otimes v_+ \oplus X_n \otimes v_- \oplus X_{n-1} \otimes e$$ and the differential is given by $$d(x \otimes v_\pm) = dx \otimes v_\pm, \\ d(x \otimes e) = dx \otimes e + x \otimes v_+ - x \otimes v_-.$$

And now the mapping cylinder $$Y \cup_{X \times 0} X \times I$$ is replaced by $$\operatorname{Cyl}(f) = Y \oplus_{X \otimes v_-} X \otimes I$$. It is the quotient of $$Y \oplus X \otimes I$$ where you identify $$x \otimes v_- \in X \otimes I$$ with $$f(x) \in Y$$ (recall that $$v_-$$ represents the vertex $$0 \in [0,1]$$). So concretely we get: $$\operatorname{Cyl}(f)_n = Y_n \oplus X_n \oplus X_{n-1} \\ d(y, 0, 0) = (dy, 0, 0) \\ d(0,x,0) = (0, dx, 0) \\ d(0,0,x') = (-f(x'), x', dx')$$

The first factor is the image of $$X \otimes v_-$$, which is identified with $$Y$$. The second factor is $$X \otimes v_+$$, and the last part is $$X \otimes e$$.

Now to get the mapping cone from the mapping cylinder, in algebraic topology you collapse $$X \times 1$$. The $$X \times 1$$ part in homological algebra corresponds to the middle $$X_n$$ (really $$X_n \otimes v_+$$) in $$\operatorname{Cyl}(f)_n$$, so just quotient out by this ideal to get $$\operatorname{Cone}(f)_n = Y_n \oplus X_{n-1}\\ d(y,0) = (dy, 0) \\ d(0,x') = (-f(x'), dx)$$ And this is exactly the definition of the mapping cone. There are various way to get to this result in a systematic manner. For example you can put what is called a model structure on the category of chain complexes, and then the mapping cone of $$f$$ becomes its homotopy cokernel. Or you can put a triangulated structure on it (though that's a bit circular, since you need to know what the mapping cone is to get the triangulated structure).

PS: A lot of things that are true in algebraic topology are also true in homological algebra. For example, if you have $$A \subset X$$, you can consider the cone on $$A$$ to get $$X \cup CA$$, then you can cone $$X$$ inside it to get $$(X \cup CA) \cup CX$$, and this is homotopy equivalent to the suspension $$\Sigma A$$ (the beginning of the Puppe sequence). Well, in homological algebra it's exactly the same: say you have a subcomplex $$i : A_* \to X_*$$, you can take the cone $$\operatorname{Cone}(i)$$, of which $$X_*$$ is a subcomplex; if you then take the cone of this inclusion, you get a complex homotopy equivalent to the suspension (shift in degree) of $$A_*$$. This is because all this can be encoded in the triangulated structure of chain complexes! 

• Wow this was great. All the stuff Im learning makes more sense now. – Abellan Feb 26 '16 at 9:33
• can you make a few points more precise for me? 1. What is the precise way you mentiond above of I_* to be [0,1]? 2. why is X_* \otimes I isomorphic to the shifted direct sum? and 3. why does the twist of the differential represent the identification? bg, Johannes – Johannes Mar 14 '17 at 13:16
• i can answer my second question on my own: by the definition of the tensor product for chain complexes – Johannes Mar 14 '17 at 16:47
• @Johannes While trying to answer your question I realized I had made a mistake while writing this answer... This should be more coherent now. I've added something about your first question, but it's not easy to answer if you don't know what model categories are. Alternatively, you can view $[0,1] = \Delta^1$ as a simplicial set, then $I_*$ is the reduced chain complex associated to $\Delta^1$. – Najib Idrissi Mar 16 '17 at 12:36
• @NajibIdrissi thnx a lot. It definitely explains enough for me (even w/o knowledge of model categories). One last Question: Why does the first differential on the tensor is the identity? – Johannes Mar 18 '17 at 15:54