Do we distinguish two singular simplices if they have different vertex orders?

We define a $\textbf{singular $n$-simplex}$ in $X$ to be a continuous map $\sigma:\Delta^n\to X$ where $\Delta^n$ is the standard $n$-simplex. Now, as an example, Let $X$ be a singleton $\{p\}$. Then is the number of $1$-simplices in $X$ one or two? I think it could be one because there is only one continuous map on $\Delta^1$ into $X$. But since $\Delta^1$ has two orders $[v_0,v_1]$ and $[v_1,v_0]$, I think the answer could be two (namely, $[v_0,v_1]\to X$ and $[v_1,v_0]\to X$).

Solution 1:

The answer is one. There is only one continuous map here, so the answer is one. Actually, both the maps that you are mentioning are the same map.

Solution 2:

This is an interesting question.

You got two answers conforming that there is only one continuous map on $\Delta^1$ into $X$, and of course this is correct. The standard $n$-simplex $\Delta^n$ is a topological space and nothing else, therefore the continuous maps living on $\Delta^n$ do not depend on a vertex ordering.

But if we consider the complete singular complex $C_*(X) = (C_n(X), \partial_n )$ of $X$, we see that the boundaries $\partial_n : C_n(X) \to C_{n-1}(X)$ depend on a particular vertex ordering of $\Delta^n$. In fact, the standard $n$-simplex is the convex hull of the $n+1$ standard basis vectors $e^{n+1}_0,\ldots, e^{n+1}_n$ of $\mathbb R^{n+1}$. The indices $i = 0,\ldots,n$ give a natural vertex ordering of $\Delta^n$. This allows to define $n+1$ face-emdeddings $\delta_i : \Delta^{n-1} \to \Delta^n$, $i = 0,\ldots,n$, by mapping $e^n_j$ to $e^{n+1}_j$ for $i < j$ and to $e^{n+1}_{j+1}$ for $j \ge i$. Then we define $$\partial_n(\sigma) = \sum_{i=0}^n (-1)^i \sigma \delta_i .$$

You see that the face-emdeddings $\delta_i$ do depend on the vertex ordering. One could consider "ordered singular simplices" in form of pairs $(\sigma : \Delta^n \to X, o_n)$ with a vertex ordering $o_n$ of $\Delta^n$. This can be described in form of a sequence $e^{n+1}_{i_0},\ldots, e^{n+1}_{i_n}$. Any such ordering induces the following vertex ordering $\partial o_n$ of $\Delta^{n-1}$: Letting $i_k = n$, we take $e^n_{i_0}, \ldots, e^n_{i_{k-1}},e^n_{i_{k+1}},\ldots e^n_{i_n}$. Moreover, one could define face embeddings $ (\Delta^{n-1},\partial o_n) \to (\Delta^n, o_n)$ which depend on $o_n$. This would produce a variant of the singular complex. I am not going to further explore this construction, but I am sure that the resulting homology groups agree with the usual singular homology groups.

In other words: You get a more complicated construction, but do not have a benefit.

Solution 3:

By definition, the "standard $n$-simplex" is one specific topological space (usually defined as $\{(x_0,\dots,x_n)\in[0,1]^{n+1}:\sum x_i=1\}$). So, a singular $n$-simplex is literally just a map from this one specific topological space to $X$. There is no other data involved--we don't pick an ordering of its vertices. (Instead, just as the standard $n$-simplex is a specific topological space, it also has a specific canonical ordering of its vertices, given by the ordering of the coordinates in $[0,1]^{n+1}$.) So, there is just one singular $n$-simplex in a singleton space.