Hatcher's book says 'A singular n-simplex in a space is by definition just a map $\sigma:\Delta^n\rightarrow X$'. And $C_n(X)$ be the free abelian group with basis the set of singular n-simplices in $X$.

So does he mean $C_n(X)$ is the... of all singular n-simplices in $X$?

And does he identify those 'same maps with different domain'? Because I see proposition $2.8$:

If $X$ is a single point, then $H_n(X)=0$ for $n>0$ and $H_0(X)=Z$.

The proof of this says:

In this case there is a unique singular n-simplex $\sigma_n$ for each n, and $\partial(\sigma_n)=\sum(-1)^i\sigma_{n-1}$, which is $0$ for n odd and $\sigma_{n-1}$ for n even...

But by definition of the boundary map: $\partial_n(\sigma)= \sum(-1)^i \sigma|[v_0,v_1...\widehat{v_i},...v_n]$. But different i deleted gives different domain, although they all map to a single point. This is what I mean 'same maps with different domain'.

Thanks for help!

  • $\begingroup$ you precompose restriction with some map from delta^n-1 $\endgroup$
    – bart
    Mar 27 '20 at 9:12
  • $\begingroup$ @bart Can you explain more? I don't understand your comment.. I'm new to the homology theory... $\endgroup$
    – ZWJ
    Mar 27 '20 at 9:24
  • 3
    $\begingroup$ sure, so in your boundary operator in the last paragraph, where you write the restriction map, thats actually the restriction precomposed with a map $\Delta^{n-1}\rightarrow \Delta^n$ given by mapping $v_j\rightarrow v_j$ for $j<i$ and $v_j\rightarrow v_{j+1}$ for $i<j$ (and extending linearly) $\endgroup$
    – bart
    Mar 27 '20 at 9:42

$C_n(X)$ is the set of all formal sums of $n-$simplices in $X$, an element of $C_n(X)$ is given by a finite sum $\sum n_i \phi_i$ for some $\phi_i : \Delta^n \rightarrow X$ and $n_i \in \mathbb Z$. This is just a formal sum. And yes he means that $C_n(X)$ is the free abelian group generated by the set of $\textbf{all}$ n-simplices.

And that definition of the boundary map is really, really misleading, I could 100% see how you would think that deleting different $i$ give different domains. And according to that definition that is actually true. A better definition is $\partial \sigma = \sum (-1)^i \sigma \circ d_i$ where $\sigma : \Delta^n \rightarrow X$ and

$$d_i: \Delta^{n-1} \rightarrow \Delta^n : (x_0,...,x_{n-1}) \mapsto (x_0,...,x_{i-1}, 0,x_{i},...,x_{n-1})$$

then the resulting maps have the same domain and the resulting formal sum is in $C_{n-1}(X)$.

The map $d_i$ is the same as the map that @bart described in his comment by the way.


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