# How is the homomorphism $b: LC_n(Y) \to LC_{n+1}(Y)$ where $b[w_0\cdots w_n] \mapsto [bw_0\cdots w_n]$ well-defined?

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 $$C_n(Y)$$.
• 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? – Al Jebr Mar 6 at 20:31
• 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. – William Mar 6 at 21:37
• But it is possible for the barycenter of $\Delta^2$ to map to the same point as one of the vertices. No? – Al Jebr Mar 6 at 22:50
• 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 $\sigma(v_2)$. – William Mar 6 at 23:24