Algebraic Topology Artin

Question

I am reading algebraic topology by Artin, and he gives the following definition. Let $X$ be a topological space. We define a homomorphism $sd_X: S_p(X) \to S_p(X)$ by induction. If $T: \Delta_0 \to X$ is a singular 0-simplex, we define $sd_XT = T$. Now suppose $sd_X$ is defined in dimensions less than $p$. If $i_p: \Delta_p \to \Delta_p$ is the identity map, let $\hat{\Delta}_p$ denote the barycenter of $\Delta_p$ and define $sd_{\Delta_p}i_p = (-1)^p[sd_{\Delta_p}(\partial i_p), \hat{\Delta}_p]$. Then define $sd_XT = T_\#(sd_{\Delta_p} i_p$.

My question is: what does $sd_{\Delta_p}(\partial i_p)$ mean?

score 1 · Accepted Answer · answered Mar 24 '13 at 03:47

1

$i_p$ is a singular $p$-simplex in the space $\Delta_p$, so it has a boundary $\partial i_p$, which is a $(p-1)$-chain in the same space. By induction, you already have defined the map $sd_X$ on all $(p-1)$-chains of all spaces $X$, so in particular you have it defined on $(p-1)$-chains of the space $\Delta_p$, and you can then compute $sd_{\Delta_p}(\partial i_p)$.

I remember staring at that in that book for a long time! :-)

answered Mar 24 '13 at 03:47

Mariano Suárez-Álvarez

135,076

Thanks that helps. However, I was wondering what $sd_{\Delta_n}(\partial i_n)$ would equal? Is this computable? I can't seem to compute it. – Susan Mar 24 '13 at 04:31
Of course it is computable :-) It is defined by a recursive procedure: follow it! – Mariano Suárez-Álvarez Mar 24 '13 at 04:57

Algebraic Topology Artin

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