I came across this definition that a singular $n$-simplex in a topological space $X$ is a continuous map $\sigma\colon \Delta^n \to X$. Using this definition a few examples were put forward: a singular $0$-simplex is a point, a singular $1$-simplex is an essential path in $X$. Now, I am aware that a $0$-simplex is a point and a $1$-simplex is a line. Can someone explain the similarities of these two examples?

  • 1
    $\begingroup$ Careful how you use the word 'essential'. $\endgroup$
    – Tyrone
    May 31, 2020 at 13:38
  • $\begingroup$ @Tyrone Actually I did not put it that way. I found it that way. The example further goes on to show that if we define a function f by $f(t) = \phi (1-t,t).$ Then f gives a path $f: C\to X$ from $\phi(v1)$ to $\phi(v2)$ and vice versa. This is where the word essential stems out. $\endgroup$
    – Ayan Shah
    May 31, 2020 at 13:45
  • $\begingroup$ It's usual to say that a map $f:X\rightarrow Y$ between spaces $X,Y$ is essential if it is not null homotopic. I don't think this is what you mean when you use the word. $\endgroup$
    – Tyrone
    May 31, 2020 at 13:48
  • $\begingroup$ @Tyrone Okay, so what difference does essential path create ? $\endgroup$
    – Ayan Shah
    May 31, 2020 at 14:06
  • $\begingroup$ A singular $1$-simplex is any continuous function $\Delta^1 \to X$. It could in particular be a constant function, which is trivially null-homotopic and therefore not "essential" in that sense. In fact since every simplex $\Delta^n$ is contractible every singular $n$-simplex is null-homotopic as a function, though it's possible that the image is a non-contractible subspace of $X$. $\endgroup$
    – William
    May 31, 2020 at 16:42

1 Answer 1


The reason of considering singular simplices is because simplicial homology is only defined for simplicial complexes (or more generally $\Delta$-complexes). In that context the simplices are part of the underlying structure of the object you're working with, and using the properties of this structure it's clear how to define the "boundary" of a simplex as a formal sum of simplices of one dimension lower (being careful to take orientations into account so that the boundary operator is a differential on the chain groups).

But say we want to define a homology theory for ALL spaces. We like the combinatorial nature of simplicial homology and we want to replicate it as much as we can in the most general context, but there are no simplices given to us. So instead, where we would normally have a set of $k$-simplices given as part of the structure, the most natural thing to do is to say that "a simplex in $X$" is just a continuous function $\Delta^k \to X$. The "singular" part comes from the fact that these don't have to be embeddings, for example this includes constant functions. The "boundary" of a singular simplex is expressed as a formal sum of singular $(k-1)$-simplicies in an analogous way, by precomposing with the various simplicial face maps $\Delta^{k-1} \to \Delta^k$. Then to construct singular homology you consider free abelian group on the set of all continuous functions $$C_k^{sing}(X) = \mathbb{Z}\langle C(\Delta^k, X) \rangle = \mathbb{Z}\langle \{ \Delta^k \to X \} \rangle$$ with the appropriate boundary operator.

For a non-trivial example consider $\Delta^1\cong [0,1] \to S^1$ given by $t\mapsto e^{2\pi i t}$. This singular $1$-simplex is actually a $1$-cycle, and generates the singular homology group $H_1^{sing}(S^1)$. In fact this is part of a general example: for all $k>0$ the quotient space $\Delta^k/\partial \Delta^k$ is homeomorphic to $S^n$, and a "quotient map" $\Delta^k \to S^k$ will generate $H^{sing}_k(S^k)$.

Note: As continuous functions all singular simplices are automatically null-homotopic because $\Delta^k$ is contractible for all $k$, but this fine in terms of homology because the homotopy relation is not used at all in the definitions of our groups.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .