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 '19 at 20:27
  • 1
    $\begingroup$ Sorry for my confusion, but what are $C_*^{\mathcal U}(X)$ and $H_p^{\mathcal U}(X)$? I am not familiar with that notation. $\endgroup$ – Arun Debray Nov 7 '19 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 '19 at 21:10
  • 2
    $\begingroup$ 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. $\endgroup$ – Charles Rezk Nov 8 '19 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. $\endgroup$ – Charles Rezk Nov 8 '19 at 17:10

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 '19 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 '19 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. $\endgroup$ – Allen Hatcher Nov 14 '19 at 17:56
  • $\begingroup$ No, if you actually do the computations by hand there won't be 28, but 16 chains. I will eventually do them again and post it in another coment on this answer. $\endgroup$ – Akerbeltz Nov 14 '19 at 18:05
  • 1
    $\begingroup$ @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. $\endgroup$ – Allen Hatcher Nov 17 '19 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.