0
$\begingroup$

Let $\Delta$ be the $n-$simplex with vertices $1,\dots,n$. I want to prove that the reduced homology of $\Delta$ is trivial without making an explicit mention to the fact that $\Delta$ is contractible.

I can use the fact that chain-homotopic chain complex maps induce the same homomorphism in homology, so I just need to find such maps and such chain-homotopy.

For a fixed vertex $x$ of $\Delta$, I've defined the inclusion map $C(\{x\})\to C(\Delta)$ and the candidate for its homotopical inverse is just the constant map which maps every vertex onto $x$. This way, the chain map is the inclusion in $C_0$, and it's $0$ in $C_i$ for $i>0$.

The composition map $q:C(\Delta)\to C(\Delta)$ is a chain map defined as follows: $q_0$ is induced by the constant map $k\mapsto x$ for every vertex $k$ of $\Delta$, and $q_i=0$ for $i\geq 1$.

I want to prove that $q$ is chain-homotopic to the identity map, so that it's proven that $C(\{x\}),C(\Delta)$ are chain-homotopically equivalent and they have the same homology. But I'm having a bit of trouble defining the chain homotopy $P:C_i(\Delta)\to C_{i+1}(\Delta)$.

Similarly to the proof of 2.10 in Hatcher's book, page 112, I want to define such a prism operator, but I get a bit confused since now I am in simplicial homology and I don't know how would I translate the homotopy into the corresponding part of the prism operator in simplicial homology.

$\endgroup$

1 Answer 1

1
$\begingroup$

In simplicial homology one constructs a contracting homotopy as follows. For a simplex $\sigma$ with vertices $i_0<i_1<\cdots<i_k$ define $P(\sigma)=\tau$ with vertices $1<i_0<\cdots<i_k$, when $i_0>1$ and $P(\sigma)=0$ otherwise. This is a homotopy between the identity map on $C(\Delta)$ and the simplicial map collapsing $\Delta$ onto the vertex $1$.

$\endgroup$
2
  • $\begingroup$ Oh, that's good. Is there a particular intuitive reasoning which helps to find it? I mean, it doesn't seem a weird map, I just want to know how you decided it was that one. $\endgroup$ Commented Sep 19, 2019 at 2:17
  • $\begingroup$ @DavidMolano I've seen it before.... I suppose it's an obvious thing to try, and it works. $\endgroup$ Commented Sep 19, 2019 at 5:26

You must log in to answer this question.

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