TITLE: Definition of $T$ operator in proof of Excision Theorem on Homology Theory
QUESTION [13 upvotes]: I had some problems with the proof of Excision Theorem.
I use two books at the same time, one is (Algebraic Topology, Hatcher) and the other one is "An Introduction of Algebraic Topology", Rotman.
In the first book the main work is done in Proposition $2.21$ in four steps. At the second step is defined an operator $T$, that work as chain homotopy between the identity and the barycentric subdivision operator $S$ on $LC_n(Y)$, with $Y$ convex. Hatcher defines $T$ operator inductively as $ \ T_n\lambda=b_{\lambda}(\lambda-T_{n-1}\partial_n \lambda)$. Is proved that the formula $\partial_{n+1} T_n +T_{n-1} \partial_n=1-S$ holds at degree $n$ using the same formula at degree $n-1$.
The same fact in Rotman is proved in Lemma $6.13$ using $T$ operator defined as $T_n \gamma =b.(\gamma -Sd \gamma -T_{n-1}\partial_n \gamma)$, where $b.$ is the cone operator $b_{\lambda}$ used by Hatcher, and $Sd_n$ operator is $S$ in the notation above. Also in this case the formula $\partial_{n+1} T_n +T_{n-1} \partial_n=1-Sd_n$ holds using the properties of cone operator and the formula 
$\partial_n(\gamma -Sd_n \gamma-T_{n-1} \partial_n \gamma)=\partial_n \gamma - Sd_{n-1}\partial_n-(1-Sd_{n-1}-T_{n-2} \partial_{n-1}) \partial_n \gamma$
that holds by induction and properties of $Sd$ operator.
The two definitions of $T$ are not the same and differ by an $Sd_n$ term that appears in Rotman definition and in Hatcher doesn't.
$1)$ Can someone explain me the reasons (algebraic and geometric) of this difference?
$2)$ Hatcher and other authors say that the geometric interpretation of $T$ operator is a subdivision of $\Delta^n \times I$ obtained by joining simplices in $\Delta^n \times\{0\} \cup (\partial \Delta^n) \times I $ to the barycenter of $\Delta^n \times \{ 1\}$.
I don't understand this explanation, can someone be more clear?

REPLY [3 votes]: I am following Hatcher. I don't think answering so late makes much sense, but still I present my interpretation for your consideration. 
We need an operator T that satisfies $\partial_{n+1} T_n +T_{n-1} \partial_n=1-S$.
Now for simplicity and motivation we look at a $2$-simplex as illustrated in Hatcher. Whenever we need to define a chain homotopy like $T$ we think of a way to decompose $\Delta^n \times I$ to some $n+1$ simplices. $T$ describes this decomposition. The decomposition should be such that, boundary $\partial T \lambda$ of  the $3$-simplices when summed up in a certain way (to be thought of next) cancel at all the faces except the ones remaining in the linear combination of $2$-simplices appearing in $(1-S)\lambda-T\partial\lambda$. This hints (though vaguely) that the decomposition should give us some $3$-simplices with the edges and vertices as those appearing in the barycentric division of the $2$-simplex. Thus the decomposition (as in the illustration in Hatcher) is in some sense not that out of the blue. 
Now we try to define $T$ on $\lambda$ for the $2$-simplex case. We want $T$ to be the map which projects a particular linear combination of the $3$-simplices to the $2$-simplex  followed by composition with $\lambda$. Given the way $T\partial\lambda$ looks from Hatcher's calculation we see that $b_\lambda(\lambda - T\partial\lambda)$ is the one possible linear combination (by a brute brute force calculation), and $T$ is the map which projects this sum into the $2$-simplex followed by composition with $\lambda$ (Here $b_\lambda$ is to be identified with the barycenter of $\Delta_{n} \times \{1\}$) . But here we have assumed the definition of $T$ on $1$-simplices. In order to verify that we would have followed same the line of reasoning.We then have a form for $T$ for the $2$-simplex (verified for higher $n$, anyhow considering a 3-simplex for geometric motivation is unyielding due to obvious limitations), which works for higher n when defined as in Hatcher.
P.S. Apologies for poor presentation and vague answer