research34 min studyLesson 548 in Mathematics

Triangulated structures and mapping cones

Build a signed cone, verify its differential, and use triangles without treating them as arbitrary exact sequences.

On this page

What you’ll understand

  • Verify the cochain shift and cone signs directly.
  • Calculate the cohomology of a cone and detect a quasi-isomorphism.
  • Explain rotation, comparison, and the role of the octahedral axiom.

A mapping cone combines a map and its failure to be a cohomological equivalence into one complex. It also supplies the basic triangles used in homotopy and derived categories. Signs are structural here: changing a shift differential or the last triangle map without adjusting the rest can break compatibility with connecting maps. We use the Stacks convention throughout: the cone differential has a positive off-diagonal map, and its distinguished triangle ends with the negative projection to the shifted source.

Build the shift before the cone

For a cochain complex, the shifted term C[1]nC[1]^n is Cn+1C^{n+1} and its differential is the negative of the original differential. Given a cochain map f:C→Df:C\to D, the cone term is Dn⊕Cn+1D^n\oplus C^{n+1}. Apply the proposed differential twice: the first component contains dDf(x)−f(dCx)d_Df(x)-f(d_Cx), which vanishes by the cochain-map identity, while the remaining square terms vanish separately. This calculation both checks the sign and explains why a map of graded modules alone would not suffice.

The inclusion i:D→Cone⁡(f)i:D\to\operatorname{Cone}(f) and projection p:Cone⁡(f)→C[1]p:\operatorname{Cone}(f)\to C[1] are cochain maps with these conventions. To match the positive connecting maps obtained by lifting and differentiating a short exact sequence, our distinguished cone triangle is (f,i,−p)(f,i,-p). Other textbooks may choose a different last-map convention. Either convention can be used consistently, but borrowing one displayed arrow from the other convention without a sign translation changes the triangle.

Cone⁡(f)n=Dn⊕Cn+1,d(y,x)=(dDy+f(x),−dCx)\operatorname{Cone}(f)^n=D^n\oplus C^{n+1},\qquad d(y,x)=(d_Dy+f(x),-d_Cx)
The shifted source has negative differential; the off-diagonal term is the given cochain map.↔ Scroll wide equations horizontally.

Triangles organize the missing information

In the homotopy category, distinguished triangles are those isomorphic to the specified signed cone triangles; they pass to distinguished triangles after derived localization. Their cohomology gives a long exact sequence. A map is a quasi-isomorphism exactly when its cone is acyclic, by exactness of that sequence. Being acyclic need not make the cone contractible, so the same criterion does not characterize homotopy equivalences without replacing acyclicity by the stronger chain-level condition.

For a degreewise short exact sequence of complexes, its associated triangle is distinguished in the derived category. A degreewise split sequence also has a corresponding triangle in the homotopy category, because the chosen graded splitting exposes the off-diagonal differential. A general nonsplit short exact sequence need not give that homotopy-category conclusion directly. Distinguishing these statements avoids silently assuming that every quotient admits a compatible splitting before the derived category has been formed.

C→fD→iCone⁡(f)→−pC[1],D→iCone⁡(f)→−pC[1]→−f[1]D[1]C\xrightarrow{f}D\xrightarrow{i}\operatorname{Cone}(f)\xrightarrow{-p}C[1],\qquad D\xrightarrow{i}\operatorname{Cone}(f)\xrightarrow{-p}C[1]\xrightarrow{-f[1]}D[1]
Rotation preserves distinguishedness and places a minus sign on the shifted first map.↔ Scroll wide equations horizontally.

Axioms constrain more than adjacent composites

The triangle axioms include completing every morphism to a distinguished triangle, closure under the signed rotation, completing a compatible pair of maps to a map of triangles, and an octahedral compatibility for composable morphisms. These are structural requirements, not consequences of adjacent composites being zero. The octahedral axiom relates cones of two maps and their composite, expressing how successive failures of equivalence fit together. Proving the full axioms is beyond this lesson; the cone construction provides their concrete source.

At the level of strict cochain maps and strictly commuting squares, cone maps can be constructed by block matrices. For a square commuting only up to homotopy, a chosen homotopy contributes an extra block; changing that choice can change the resulting map. Thus a triangulated category does not, from its axioms alone, provide a canonical functorial cone for every arrow. Keeping the explicit complex or a richer enhancement retains choices that the triangulated description intentionally suppresses.

The cone of multiplication by seven

Let C=D=Z[0]C=D=\mathbb Z[0] and let ff be multiplication by seven. Determine the cone, its cohomology, and the last map of the distinguished triangle in the convention fixed above.

  1. The only cone terms are C0=ZC^0=\mathbb Z in cone degree minus one and D0=ZD^0=\mathbb Z in cone degree zero. The off-diagonal entry makes the differential multiplication by seven.

  2. Its degree-minus-one kernel is zero, while its degree-zero cokernel is Z/7Z\mathbb Z/7\mathbb Z. All other terms vanish. Hence the cone is not acyclic and multiplication by seven is not a quasi-isomorphism.

  3. The projection to C[1]C[1] is the identity on the degree-minus-one source summand. The distinguished triangle uses its negative, so the final map is negative identity in that degree and zero elsewhere. This sign is separate from the positive-seven cone differential.

The result

The cone is a free resolution of the integers modulo seven placed in nonpositive degrees. Its surviving quotient measures exactly why the original map fails to be a cohomological equivalence.

Explore the model

Interactive lab

Interactive lab · matrix

An invertible two-term cone differential

Pause/Reset, then set both diagonal scale to 1 and shear to 2. Predict determinant one, zero kernel, and zero cokernel over the reals. Compare the selected-matrix readouts with the deformed square. Reset and set both scale to 0.25 and shear to 0; the determinant changes but invertibility persists. Playback interpolates the picture and does not calculate cone cohomology.

An invertible two-term cone differentialThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA cone of a map concentrated in degree zero is a two-term complex; an invertible real differential makes that cone contractible. The lab cannot model integer torsion.

Check your understanding

With a cochain cone differential d(y,x)=(d y+f x,−d x), which stated convention is used here?

Try it yourself

For the identity map of Z[0]\mathbb Z[0], construct an explicit contraction of its cone. Then compute the cone of the zero map of the same source and target and explain why it is not acyclic.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Both cones occupy degrees minus one and zero. Determine whether their single differential is identity or zero before choosing a degree-lowering homotopy.

Compare with the solution

For the identity map, the cone differential from degree minus one to zero is identity. Taking h0h^0 from degree zero to minus one to be identity gives dh+hd=1dh+hd=1 in both degrees, so the cone is contractible. For the zero map, the differential is zero, leaving Z\mathbb Z in each of those two cohomological degrees. It is Z[0]⊕Z[1]\mathbb Z[0]\oplus\mathbb Z[1] as a complex and is not acyclic. Thus the cone detects the actual map, even when source and target are unchanged.

Take it with you

  • Check the shifted differential and the cone's final arrow together.
  • An acyclic cone detects a quasi-isomorphism, not automatically a homotopy inverse.
  • Triangle axioms organize comparison and composition without making cones canonically functorial.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.