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 isomorphism in singular homology

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

For all $p\geq 0$. In both references, a chain homotopy $h:C_p(X)\longrightarrow C_{p+1}(X)$ between the barycentric subdivision operador and the identity map is given by

If $p=0$, $h$ is the zero homomorphism. If we have defined $h$ up to some $p\in\mathbb{N}$, and $\sigma$ is a $p-$simplex in $X$, then

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

Where $b_p$ is the barycentre of the standard $p-$simplex $\Delta_p$ and $*$ is the cone operator. We then extend $h$ linearly to singular chains: $h\big(\sum_{i\in I}n_i\sigma_i\big)=\sum_{i\in I}n_ih\sigma_i$

In contrast with the chain homotopy that appears in the proof of the homotopy axiom, this is really less intuitive, and relies heavily on the equation

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

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.

  • $\begingroup$ Perhaps, this is about the acyclic carriers. $\endgroup$
    – Wlod AA
    Nov 7, 2019 at 20:27
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    $\begingroup$ Sorry for my confusion, but what are $C_*^{\mathcal U}(X)$ and $H_p^{\mathcal U}(X)$? I am not familiar with that notation. $\endgroup$ Nov 7, 2019 at 21:02
  • $\begingroup$ If $\mathcal{U}$ is an oper cover of a topological space $X$, $C_\bullet^\mathcal{U}(X)$ represents the chain complex of \mathcal{U}-small chains in $X$, whose elements are chains such that the image of each of its simplices is contained in some element of the cover $\mathcal{U}$. $H_p^\mathcal{U}(X)$ is just the homology of such complex. $\endgroup$
    – Akerbeltz
    Nov 7, 2019 at 21:10
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    $\begingroup$ I recommend computing $h$ for a single simplex $\sigma$ of low dimension (e.g., 0, 1, 2). This might give you a better idea of what is goin on. $\endgroup$ Nov 8, 2019 at 17:09
  • $\begingroup$ In particular, the formula for the boundary of the cone is just saying: the boundary of a cone is the union of the "hat" part of the cone and the "base" of the cone. $\endgroup$ Nov 8, 2019 at 17:10

1 Answer 1


When thinking about chain homotopies in a setting involving simplices it can be helpful to consider the product $\Delta^p\times I$ where $\Delta^p$ is a $p$-simplex and $I=[0,1]$. The formula $h\sigma=\sigma_{\sharp} b_p \ast(i_p-si_p-h\partial i_p)$ corresponds to a certain inductively defined subdivision of $\Delta^p\times I$ obtained by coning off a subdivision of $\Delta^p\times\partial I \cup \partial \Delta^p \times I$ to a point in the interior of $\Delta^p\times I$. The subdivision of $\Delta^p\times\partial I \cup \partial \Delta^p \times I$ is $\Delta^p$ itself (unsubdivided) on $\Delta^p\times \{0\}$ and the barycentric subdivision of $\Delta^p$ on $\Delta^p\times\{1\}$. These are the terms $i_p$ and $si_p$ in the formula. On $\partial \Delta^p\times I$ one uses the subdivision given by induction. This is the term $h\partial i_p$. The term $\sigma_{\sharp} b_p$ corresponds to the point in the interior of $\Delta^p\times I$ that one cones off to, with the symbol $\ast$ denoting the coning operation.

What is perhaps most puzzling is that the formula says nothing about taking the product with $I$, but this is because in reality one takes the subdivision of $\Delta^p\times I$ and projects it to $\Delta^p$ before applying the map $\sigma$, whose domain is $\Delta^p$ rather than $\Delta^p \times I$.

I have seen this method of subdividing $\Delta^p\times I$ in several books when they are developing homology theory, but it is more complicated than necessary. A simpler subdivision that suffices is to cone off a subdivision of $\Delta^p\times \{0\} \cup \partial \Delta^p \times I$ to the barycenter of $\Delta^p\times\{1\}$, where $\Delta^p\times \{0\}$ is unsubdivided and $\partial \Delta^p \times I$ has the subdivision given inductively. On $\Delta^p\times \{1\}$ this gives just the usual barycentric subdivision, which is also defined inductively. There is a picture of this subdivision of $\Delta^p\times I$ in the case $p=2$ on page 122 of my algebraic topology book. Perhaps other books such as the Lee book you mention don't give a picture because the picture would be more complicated for the more complicated subdivision. An advantage of the simpler subdivision is that the formula for $h\sigma$ becomes just $\sigma_{\sharp} b_p \ast(i_p-h\partial i_p)$, without the term $si_p$.

The more complicated formula is given in the classic book of Eilenberg and Steenrod (page 197) without pictures or explanation. Perhaps other books are just following suit.

  • $\begingroup$ Yes, there was a small typo in the original question. This answer is quite fulfilling, thanks. $\endgroup$
    – Akerbeltz
    Nov 11, 2019 at 21:18
  • $\begingroup$ If anyone is wondering how the cone of such subdivision of $\Delta_p\times\partial I\cup\partial\Delta_p\times I$ looks like, here is a representation of the case $p=2$: geogebra.org/3d/k8nr2wzs $\endgroup$
    – Akerbeltz
    Nov 13, 2019 at 3:20
  • $\begingroup$ @Akerbeltz: Nice graphic at the link you provided, but something seems to be missing. There are 16 tetrahedra shown but there should be 28. $\endgroup$ Nov 14, 2019 at 17:56
  • $\begingroup$ @Akerbeltz: If you form the cone on a triangulation of $\Delta_p\times \partial I \cup \partial \Delta_p \times I$ then the result is a triangulation of $\Delta_p\times I$. In your picture with 16 tetrahedra the underlying topological space is not homeomorphic to $\Delta_2\times I$ so it cannot give a triangulation of $\Delta_2\times I$. (A horizontal plane halfway between the top and bottom of your figure intersects the figure in something one-dimensional instead of two-dimensional.) $\endgroup$ Nov 15, 2019 at 15:22
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    $\begingroup$ @Akerbeltz: I think you are understanding this correctly. The twelve missing tetrahedra are grouped into six pairs with the two tetrahedra of each pair interchanged by reflecting the $I$ factor of $\Delta_2\times I$ across its midpoint. This means that after projecting $\Delta_2\times I$ onto $\Delta_2$ the two tetrahedra in each pair cancel algebraically. The inductive construction implies that there is a similar cancelation in all dimensions. $\endgroup$ Nov 17, 2019 at 13:45

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