# Chain homotopy on linear chains: confusion from Hatcher's book

I am following the book of Hatcher, see page 121-122 for notations.

Define $T:LC_n(Y)\rightarrow LC_{n+1}(Y)$ as follows inductively: for $n=-1$ it is zero map. For $n\geq 0$,
$$T(\lambda):=b_{\lambda}(\lambda-T\partial\lambda).$$

Question. Consider $n=0$. If $\lambda:\Delta^0\rightarrow Y$, $\lambda(v_0)=w_0$ then $\lambda=[w_0]$ (by notation) and so $$T(\lambda)=b_{\lambda}(\lambda-T\partial\lambda)=b_{\lambda}([w_0]-T\partial[w_0])=b_{\lambda}([w_0]-T[\phi])=b_{\lambda}[w_0]=[w_0,w_0];$$ is this correct? If yes, move to $n=1$, for $\mu:\Delta^1\rightarrow Y$, $\mu(v_0)=w_0, \mu(v_1)=w_1$, we have $$T(\mu)=b_{\mu}(\mu-T\partial\mu).$$ Here $T\partial\mu=T([w_1]-[w_0])=T[w_1]-T[w_0]=[w_1,w_1]-[w_0,w_0]$ is this correct? Therefore $$T(\mu)=b_{\mu}([w_0,w_1]-[w_1,w_1]+[w_0,w_0])=[b,w_0,w_1]-[b,w_1,w_1]+[b,w_0,w_0].$$ Here $b$ is image of barycenter of $[v_0,v_1]$ under $\mu$ in $Y$. This last calculation I am not realizing well geometrically as Hatcher explains on page 122; I feel I am wrong somewhere, or misunderstanding something. Can one explain this?

Notations

• $\Delta^n=[v_1,v_1,\cdots,v_n]$ is the standard $n$-simplex of dimension $n$ in Euclidean space.
• $C_n(X)$ denote the free abelian group on all continuous maps $\Delta^n\rightarrow X$.

• $LC_n(X)$ denotes the subgroup of $C_n(X)$ generated by 'linear maps' $\Delta^n\rightarrow X$.

• A 'linear map' $\lambda:\Delta^n\rightarrow X$ is also denote by $[\lambda(v_0),\cdots, \lambda(v_n)]$ (repetition is possible in images).

• For $b\in Y$, define $b:LC_n(Y)\rightarrow LC_{n+1}(Y)$ by $b([w_0,\cdots,w_n])=[b,w_0,\cdots,w_n]$.

• For $\lambda:\Delta^n\rightarrow Y$, define $S(\lambda):=b_{\lambda}(S\partial \lambda)$ with $S([\phi)]=[\phi]$ (empty simplex).

• I am studying this now and getting the same answers you have. I have the same questions as you now. Did you end up figuring it out? – Juan Lanfranco Mar 6 at 23:56
• I think the singular simplices like $[b, w_i, w_i]$ are not intuitive since they are not embeddings of $2$-simplices. But if you draw out the $n=1$ case with the points $w_0, b_\lambda, w_1$ making an $L$-shape, you can see it more clearly and it is similar to Hatcher's $n=2$ case. – Juan Lanfranco Mar 7 at 0:04