Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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 ?
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! $$ $$
