1
$\begingroup$

I am working through some examples on the homology of mapping torii in Hatcher's Algebraic Topology. One thing that is confusing me is the following:

Hatcher – Chapter 2

I don't see why the map on the zeroth homology must be zero. We know that $\deg g=-1$ and thus $g$ is homotopic to minus the identity. Why then, is the induced map $\mathbb{1}-g_*=0$ on zeroth homology?

I'm sure this is simple, but I can't see it myself.

Edit: more generally, how does a degree $n$ map act on the zeroth homology? It seems 'obvious' that it should induce the identity map, but why...

$\endgroup$
  • 3
    $\begingroup$ The induced map on zeroeth homology of a map between path-connected spaces is always the identity. Try to prove this, and then generalize to non-path-connected spaces! $\endgroup$ – user98602 May 16 '16 at 17:16
2
$\begingroup$

Recall that the $H_0$ of a path-connected space $X$ is isomorphic to $\mathbb{Z}$ via the augmentation map, which counts the sum of the coefficients of a given (class of a) $0$-chain, i.e., a formal sum of points of $X$. Therefore the generator $1 \in \mathbb{Z}$ corresponds to the class of any point $x \in X$; alternatively, if you pick $x,y \in X$, then they both belong to the same class because their difference is the boundary of a path ($1$-simplex) from one to the other.

Therefore, any continuous $g:X \to X$ induces the identity at the level of $H_0$ because $$1 \leftrightarrow [x] \mapsto [g(x)] \leftrightarrow1,$$

i.e. $1$ is mapped to $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.