Given a group G, the example 1B.7 in Hatcher's Algebraic Topology constructs a K(G, 1) space by taking the simplices $[g_0,\ldots,g_n]$ of elements of G with the faces attached in the obvious way.

My first answer is: what happens when some of the $g_i$ are the same?

Then it defines an homotopy of the complex to the vertex [e] by sliding every point in $[g_0,\ldots,g_n]$ along the segment in $[e, g_0,\ldots,g_n]$. However it clarifies that [e] is not fixed because it slides along the loop [e, e].

What happens when $g_i=e$? If it's a loop, how can it contracts to [e]?(this question probably means that the situation is not clear to me)

And finally, it seems to me that by collapsing every degenerate simplex (for example considering [e, e, a] as [e, a]) we obtain a much simpler proof, so what makes it wrong?


1 Answer 1


What happens when some of the $g_i$ are the same?

Hatcher is constructing a $\Delta$-complex, which allows repetition of vertices, unlike for simplicial complexes. For example, if $[a,a,b]$ is one of the simplicies, then its faces are $[a,a],[a,b],[a,b]$.

If it's a loop, how can it contract to [e]?

Hatcher is mentioning that $e$ slides along $[e,e]$ to point out that this contraction is not a deformation retract ($e$ is not fixed throughout the contraction). All that matters for the contraction to $e$ is that everything ends up at $e$ in some consistent, continuous way. Sure $[e,e]$ is a loop, but it contracts to $[e]$ through $[e,e,e]$, which contracts to $[e]$ through $[e,e,e,e]$, ... which contracts to $[e]$ through $[e,\dots,e]$, ... and so on.

It may seem weird throwing the problem up a dimension ad infinitum, but each $n$-simplex only travels through the $(n+1)$-skeleton, so it is OK.

It seems that by collapsing every degenerate simplex we obtain a simpler proof.

If it seems like the proof would be simpler, try writing it out. But don't confuse the complexity of the proof with the complexity of the constructed object! Sometimes introducing ridiculously large objects makes proofs simpler. (I think it does work out if you eliminate the degenerate simplices.)

One of the benefits of the construction in Hatcher is, as he says, that it is functorial. Given a homomorphism $G\to H$, there is a simple formula for the continuous map between the delta complexes. If $G\to H$ is $g\mapsto e$, then the formula says that every simplex is sent to a degenerate simplex $[e,e,\dots,e]$.

Why this construction?

You didn't ask this, but something to notice if you study group cohomology is that the move from the $[g_0,\dots,g_n]$ form to the $g_0[e|g_1|\dots|g_n]$ form is like the motion from homogeneous to inhomogenous cochains. Originally, group cohomology for a group $G$ was the cohomology of a $K(G,1)$. The chain complex of this particular simplicial structure corresponds to the standard projective resolution of $\mathbb{Z}$ as a trivial $\mathbb{Z}[G]$-module.

  • $\begingroup$ Thank for your answer. I seem to understand now that what happens to $[e,e]$ is similar to how every $S^k$ contracts in $S^{k+1}$ and that makes $S^\infty$ contractible even though none of its skeletons is $\endgroup$
    – karmalu
    Jul 7, 2017 at 9:08
  • 1
    $\begingroup$ Yes, I considered mentioning $S^\infty$ but didn't want to explain the analogy --- I'm happy to see I didn't need to! $\endgroup$ Jul 7, 2017 at 18:27

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