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.



Geometric intuition behind this chain homotopy
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.


