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I'm confused about part of the proof of Proposition 2.7 in Hatcher.

If $X$ is nonempty and path-connected, then $H_0(X)\simeq\mathbb{Z}$. Hence for any space $X$, $H_0(X)$ is a direct sum of $\mathbb{Z}$s, one for each path-component of $X$.

Proof: By definition, $H_0(X)=C_0(X)/\operatorname{Im}\partial_1$. Define a homomorphism $\epsilon\colon C_0(X)\to\mathbb{Z}$ by $\epsilon(\sum_i n_i\sigma_i)=\sum_i n_i$. This is surjective, so the proof proceeds to show $\operatorname{Im}\partial_1=\ker\epsilon$.

To show $\ker\epsilon\subset\operatorname{Im}\partial_1$, suppose $\epsilon(\sum_i n_i\sigma_i)=0$, so $\sum_i n_i=0$. The $\sigma_i$ are singular $0$-simplicies, which are simply points of $X$. Choose a path $\tau_i\colon I\to X$ from a basepoint $x_0$ to $\sigma_i(v_0)$, and let $\sigma_0$ be the singular $0$-simplex with image $x_0$. We can view $\tau_i$ as a singular $1$-simplex $\tau_i\colon [v_0,v_1]\to X$ and then we have $\partial\tau_i=\sigma_i-\sigma_0$.

This final sentence is confusing to me. I know we can think of $1$-simplicies as line segments, so it sort of makes sense to think of $\tau_i$ as a singular $1$-simplex. But I thought then $\partial\tau_i=\tau_i|[v_1]-\tau_i|[v_0]$, why does Hatcher say $\partial\tau_i=\sigma_i-\sigma_0$?

Hailie Mathieson
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2 Answers2

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You are almost all the way there. You have $\partial \tau_i=\tau_i|[v_1]-\tau_i|[v_0]$, but $\tau_i|[v_j]$ is just mapping $v_j$ to a point of $X$, i.e. it is a zero simplex. By construction, it coincides with $\sigma_i(v_j)$.

Brian Klatt
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  • Sorry, I'm confused about what the construction is. First, when viewing $\tau_i\colon I\to X$, this means $\tau_i(0)=x_0$ and $\tau_i(1)=\sigma_i(v_0)$ right? But then Hatcher changes the domain, does this mean now they we identify $v_0$ as the starting point of $I$, and $v_1$ as the end point of $I$, so that $\tau_i|[v_1]=\tau(1)=\sigma_i(v_0)$ and $\tau_i|[v_0]=\tau_i(0)=x_0$? Because I don't see how they coincide. – Hailie Mathieson Apr 04 '14 at 06:10
  • @hmlll Yes, that's correct. The interval is homeomorphic to the 1-simplex so we haven't really changed anything, and this homeomorphism identifies endpoints of the interval with the vertices of the 1-simplex. –  Apr 04 '14 at 06:16
  • @Mike Thanks. May I ask why does $\sigma_i(v_0)-x_0=\sigma_i-\sigma_0$? Is it because we identify $\sigma_i$ with its image, $\sigma_i(v_0)$, and $\sigma_0$ with its image $x_0$? So these two formal differences are "equal"? – Hailie Mathieson Apr 04 '14 at 06:18
  • When you write $x_0$ above what you really must mean is a map from a point to $x_0$. In homology you're always dealing with groups formed by formally adding maps into your space. In this sense, $x_0$ and $\sigma_0$ are equal. – Brian Klatt Apr 04 '14 at 06:21
  • @hmlll Yeah, as Brian says, Hatcher is identifying $x_0$ with $\sigma_0$ by this point for notational purposes. But $\sigma_0$ is a map from a 1-point space, and $\tau_i$ restricted to $v_0$ is a map from a 1-point space with the same image. They're not "equal", they're equal. –  Apr 04 '14 at 06:28
  • @Mike Formally, wouldn't set equality of functions require that the domains be equal? Here isn't $[v_0]$ the domain of $\sigma_i$ but $\tau_i|[v_1]$ has domain $[v_1]$? I don't see the domain of $\sigma_0$ specified, only that it maps to $x_0\in X$. – Hailie Mathieson Apr 04 '14 at 06:35
  • @hmIII I guess, sure. Normally if the domains are homeomorphic (and you can pass from one function to the other via this homeomorphism) you consider the functions the same. I would especially do this in this case, since every one-point topological spaces is homeomorphic in the most trivial way possible. –  Apr 04 '14 at 06:38
  • @Mike That's certainly clear! Thanks for your comments. – Hailie Mathieson Apr 04 '14 at 06:40
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Because $\tau_i| [v_1]$ is $\tau_i$ evaluated at $v_1$, which is just $\sigma_1$.

arando
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