This is on page $121$ of Hatcher's Algebraic Topology.

$Y$ is a convex subset in some Euclidean space. The linear maps $\Delta^n \to Y$ generate the subgroup of linear $n$-chains on $Y$, $LC_n(Y) \le C_n(Y)$.

For each $b \in Y$, a homomorphism is defined by taking a linear $n$-chain on $Y$ to a linear $(n+1)$-chain on $Y$, i.e., $b:LC_n(Y) \to LC_{n+1}(Y)$, via $[w_0\cdots w_n] \mapsto [bw_0\cdots w_n]$.

However, what if $b$ is a vertex in an $n$-simplex? We have $b([bw_1\cdots w_n])=[bbw_1\cdots w_n]$.

Is this well-defined? What is $[bbw_1\cdots w_n]$?

Hatcher also writes:

Geometrically, the homomorphism $b$ can be regarded as a cone operator, sending a linear chain to the cone having the linear chain as the base of the cone and the point b as the tip of the cone.


A linear map $\Delta^n \to Y$ doesn't have to be injective. If $b$ is already contained in an elementary chain $[w_0,\dots, w_n]$ then $b([w_0,\dots, w_n]) = [b, w_0, \dots , w_n]$ is a degenerate simplex. This is allowed in the definition of singular chains as well, for example for every $n$ the constant singular chain $\Delta^n \to *$ is a valid element of $C_n(Y)$.

For the "geometrically" part, if we have a $d$-chain $c = \sum_{i=0}^k a_i\sigma_i$ where $a_i \in \mathbb{Z}$ and $\sigma_i = [\sigma_{i,0},\dots, \sigma_{i, d}]$ is an elementary linear simplex then $$ b(c) = \sum_{i=0}^ka_i[b, \sigma_{i,0},\dots, \sigma_{i, d}]$$ or in other words, for each elementary simplex in $c$ we are taking the cone with $b$ as the cone point (the operation of "taking the cone" might leave the image of the simplex preserved, as in the first paragraph), and then we extend linearly to $C_n(Y)$.

  • $\begingroup$ I see. So, suppose we have a linear singular $2$-simplex $\sigma: \Delta^2 \to Y$. Let us label the $2$-simplex $[v_0v_1v_2]$. Suppose $\sigma(v_0)=\sigma(v_1)$, but $\sigma(v_2) \ne \sigma(v_0)$. What would be the barycenter of the image? $\endgroup$ – Al Jebr Mar 6 at 20:31
  • $\begingroup$ Not sure if my previous question is well posed. I guess maybe: what would be the image of the barycenter?? $\endgroup$ – Al Jebr Mar 6 at 20:45
  • $\begingroup$ For your linear simplex $\sigma$, the image will be the straight line from $\sigma(v_0)$ to $\sigma(v_2)$, and the image of the barycenter of $\Delta^2$ will be the midpoint of that line. $\endgroup$ – William Mar 6 at 21:37
  • $\begingroup$ But it is possible for the barycenter of $\Delta^2$ to map to the same point as one of the vertices. No? $\endgroup$ – Al Jebr Mar 6 at 22:50
  • $\begingroup$ Yes it is possible. For example consider the linear $2$-simplex $\sigma$ where $\sigma(v_2)$ is mapped to the midpoint of $\sigma(v_0)$ and $\sigma(v_1)$, then you can compute that $\sigma$ maps the barycentre of $\Delta^2$ to $\sigma(v_2)$. $\endgroup$ – William Mar 6 at 23:24

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