Warning: using Lee's Introduction to Topological Manifols notation in what follows.

My question has to do with the chain homotopy that appears in Lee's Introduction to Topological Manifols and Rotman's Introduction to Algebraic Topology proofs that the inclusion

$$C_\bullet^\mathcal{U}(X)\hookrightarrow C_\bullet(X)$$

Induces an isomorfism in singular homology

$$H_p^\mathcal{U}(X)\cong H_p(X)$$

For all $p\geq 0$. First, a series of definitions

If $\alpha=A(v_0,\dots,v_p)$ is an affine singular $p-$simplex in some convex set $K\subseteq\mathbb{R}^n$ and $w$ is any point in $K$, we define an affine singular $(p+1)-$simplex $w*\alpha$ called the cone on $\alpha$ from $w$ by:


And we extend this operator to affine chains by linearity: $w*\big(\sum_{i\in I}n_i\alpha_i\big)=\sum_{i\in I}n_i(w*\alpha_i)$

An important formula is proved in both references, regarding the relationship between the boundary and cone operators

If $c$ is an affine chain, then

$$\partial(w*c)=c-w*\partial c$$

It is obvious that if $c$ is a cycle, the last formula becomes $\partial(w*c)=c$, and Rotman calls this an intrgration formula.

Afterwards, both authors define an operator $s$ that sends affine $p-$chains into affine $p-$chains, called the barycentric subdivision operator. This is done by induction:

For $p=0$, $s=\text{Id}$

Suppose that $s$ has been defined for some $p\in\mathbb{N}$. Then, for any affine $(p+1)-$simplex $\alpha:\Delta_p\longrightarrow\mathbb{R}^n$ we set


Where $b_p$ is the barycentre of the standard singular simplex $\Delta_p$, and we extend this operator to affine chains by linearity: $s\big(\sum_{i\in I}n_i\alpha_i\big)=\sum_{i\in I}n_is\alpha_i$

Now, to extend this operator to arbitrary singular chains, note that if $\sigma$ is a singular $p-$simplex in any space $X$, then $\sigma=\sigma_{\#}i_p$, where $i_p:\Delta_p\longrightarrow\Delta_p$ is the identity map considered as an affine singular $p-$simplex in $\Delta_p$, and $\sigma_\#:C_\bullet(\Delta_p)\longrightarrow C_\bullet(X)$ is the chain map obtained from the continuous map $\sigma$.

So the idea is that if we have a singular chain $c$, then applying sucesively the subdivision operator, we will obtain a chain homologous to $c$, but whose simplices have images all lying in elements of $\mathcal{U}$.

To prove this, we should find a chain homotopy between the identity and the subdivision operator, that is, a homomorphism $h:C_p(X)\longrightarrow C_{p+1}(X)$ such that

$$h\circ\partial+\partial\circ h=\text{Id}-s$$

But Lee and Rotman gives

$$h\sigma=\sigma_\#b_p*(i_p-si_p-h\partial i_p)$$

And extend to chains by linearity. However, I am struggling to understand what geometric intuirion is behind this intriguing formula. I tried to draw what $h$ looks like when $\sigma=\text{Id}_{\Delta_1}$, but it is really hard (impossible?) to imagine this acting on $2-$simplices

enter image description here

In contrast with the chain homotopy that appears in the proof of the homotopy axiom, this is really less intuitive, and relies heavily on what Rotman calls the integration formula.

So my questions are

  • How should we understand geometrically this map $h$? What is the geometric intuition that allows us to choose this a a good chain homotopy for our purposes?

  • How should we understand the formula $\partial(w*c)=c-w*\partial c$? What is the meaning of this equation geometrically speaking?

  • How to come up with such a map in the first place? How has this theorem developed historically?

I understand perfectly both demonstrations, since the calculations are easy to follow; I am just concerned with how this map gives no intuition at first glance about the geometry involved.

  • 1
    $\begingroup$ Your PS is not needed : this is a greatly written question : there are precise questions, there's a motivation, and you show that you tried stuff : a model question for sure ! Just a note : I don't understand your definition of cone in the beginning : what does your notation $A(w,v_0,...,v_p)$ mean ? I mean, I know what cones are so I understand the rest but not sure about this $\endgroup$ Commented Nov 7, 2019 at 20:39
  • 1
    $\begingroup$ $A(v_0,\dots,v_p):\Delta_p\longrightarrow\mathbb{R}^n$ is the only affine map that sends each $e_i=(0,\dots,1^{(i},\dots,0)$ to $v_i$, where the codomain is just a suitable Euclidean space. $\endgroup$
    – Akerbeltz
    Commented Nov 7, 2019 at 20:50
  • $\begingroup$ Ok that makes sense. For your second question, did you try to draw what $w*c$ was, and what its boundary was ? Take $c$ to be a $1$-simplex at first, and then a $2$-simplex $\endgroup$ Commented Nov 7, 2019 at 20:59
  • $\begingroup$ Yes, but I don't understand how that equation plays a role into the other stuff... $\endgroup$
    – Akerbeltz
    Commented Nov 7, 2019 at 21:24

1 Answer 1


I like to think of the formula for $h\sigma$ as a description of a subdivision of the simplicial cylinder $\sigma \times [0,1]$ as a simplicial complex.

The equation $h \circ \partial + \partial \circ h = \text{Id} - s$, when rewritten as $$\partial \circ h = \text{Id} - s - h \circ \partial $$ can be understood geometrically like this:

  1. The term $\text{Id}$ on the right hand side means that the simplex $\sigma \times 0$ on the bottom is not subdivided at all.
  2. The term $-s$ on the right hand side means that we use the barycentric subdivision for the simplex $\sigma \times 1$ on the top, with barycenter $b_p \times \{1\}$.
  3. The term $-h \circ \partial$ means that for each face of $\sigma$ of one dimension lower, the corresponding vertical side of $\sigma \times [0,1]$ is subdivided according to the inductively given formula for $h$ in one dimension lower.

And the term $\partial \circ h$ on the left hand side then gives us a clue for how to define $h$ by induction: assuming by induction that $h$ is defined in one dimension lower, use the right hande side and steps 1,2,3 to subdivide the boundary of $\sigma \times [0,1]$, and then figure out some clever way to subdivide the interior also, hopefully a way that can be described by some cute formula.

Let's do this when $\sigma$ is the 1-simplex $[0,1]$. First draw $\sigma \times [0,1] = [0,1] \times [0,1]$. The cute formula $$h\sigma = \sigma_\# b_p * (i_p - s i_p - h \partial i_p) $$ tells you to do this: draw lines going from the $b_p \times \{1\} = (1/2,1)$ point on the top side $[0,1] \times \{1\}$, to all the vertices on the other sides of the square. You'll get a triangulation of the square with three triangles: one with vertices $(0,0),(0,1),(1/2,1)$, another with vertices $(0,0),(1,0),(1/2,1)$, and the third with vertices $(1,0),(1/2,1),(1,0)$.

Now let's do this when $\sigma$ is a 2-simplex. Picture $\sigma$ in the $xy$ plane of $xyz$ space. So $\sigma \times [0,1]$ is a triangular cylinder in $xyz$ space. The bottom $\sigma \times 0$ is left unsubdivided. The top $\sigma \times 1$ is given the barycentric subdivision, with $b_p \times \{1\}$ as the barycenter. The three vertical sides are, by induction, each triangulated into three 2-simplices as just discussed above. Now connect up $b_p \times 1$ with all the vertices on the other sides. You'll get a triangulation of $\sigma \times [0,1]$ into, let's see....... 7 tetrahedra? ... Nope, 10 tetrahedra.


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