So I'm trying to understand this proof in Hatcher's Algebraic Topology.

Lemma: The composition $\Delta_n(X)\xrightarrow{\partial_n} \Delta_{n-1}(X)\xrightarrow{\partial_{n-1}}\Delta_{n-2}(X)$ is zero

where here $\partial_n$ is a boundary homomorphism $$\partial_n(\sigma_{\alpha})=\sum_i (-1)^i\sigma_{\alpha}\:\vert\:[v_0\dots,\bar v_i,\dots,v_n]$$

where $[v_0,\dots,v_n]$ is an $n$-simplex and the bar $\bar v_i$ represents removing that vertex, giving a face of of the $n$-simplex.

So here's the proof:

We have $\partial_n(\sigma)=\sum_i (-1)^i\sigma\:\vert\:[v_0\dots,\bar v_i,\dots,v_n]$, and hence

$$\partial_{n-1}\partial_n=\sum_{j<i}(-1)^i(-1)^j\sigma\:|\:[v_o,\dots,\bar v_j,\dots,\bar v_i,\dots,v_n]+\sum_{j>i}(-1)^i(-1)^{j-1}\sigma\:|\:[v_o,\dots,\bar v_i,\dots,\bar v_j,\dots,v_n]$$

And the two summations cancel because after switching $i$ and $j$ in the second sum, it becomes the negative of the first.

What I'm confused about is, why are there two summations coming from this composition of the two boundary functions? And where does the $j-1$ come from in the second sum?

My lack of understanding of this proof may be from some lack of genuine understanding of these boundary maps, but I have a limited time to learn this in and this is really stumping me.

  • $\begingroup$ Sorry to answer your question with a question, but why wouldn't there be two sums coming from the composition of two boundary maps? You apply $\partial_n$ to an $n$-simplex and you get a sum, say over $i$, of $(n-1)$-simplices. Then you apply $\partial_{n-1}$, move it into the sum via linearity, and so for each of the $(n-1)$-simplices, you get a sum, say over $j$, of $(n-2)$-simplices. Do you understand what kind of object $\Delta_n(X)$ is? $\endgroup$ – user59193 Aug 1 '15 at 0:31

When taking the second boundary map, one can not remove the $ i $th term since it was already removed by the first boundary map. So $ j \neq i $ in the sum, and so it is convenient to split up the sum as $$ \sum_{\substack{ j \\ j \neq i } } \cdot = \sum_{\substack{ j \\ j < i } } \cdot + \sum_{\substack{ j \\ j > i } } \cdot $$ Now for the signs, in the definition $ \partial_n(\sigma_{\alpha})=\sum_i (-1)^i\sigma_{\alpha}\:\vert\:[v_0\dots,\bar v_i,\dots,v_n] $, you should think of the $ (-1)^i $ as $ (-1)^{\text{number of $ v $ to the left of $ v_i $} } $. When $ j < i $, there are $ j $ terms to the left so it is $ (-1)^j $. When $ j > i $, since the $ i$ th term is already gone there are only $ j-1 $ terms to the left.

I think for these types of calculations (which happen often in algebraic topology) it is more enlightening to work through the pieces of the proof with a specific example, and even better, with a picture. Try computing $ \partial^2 [0,1,2,3] $.

  • $\begingroup$ This is an absolutely wonderful answer. Thank you so much $\endgroup$ – Alex Mathers Aug 1 '15 at 3:00

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