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The following is the exercise in Algebraic topology course. Assume that $ X$ is a finite connected graph such that no vertex is the endpoint of just one edge. Let $G$ be a group acting $X$. Then for any $g\in G$, we have

$$g_\ast : H_1(X; {\bf Z}) \rightarrow H_1(X;{\bf Z})$$

Here if $H_1(X; {\bf Z})$ is a free abelian of rank $n$, then $g_\ast \in GL(n, {\bf Z})$. And if we replace coefficient group ${\bf Z}$ by ${\bf Z}_m$ ($m> 2$) then still $g_\ast \in GL(n, {\bf Z}_m)$.

Until now I can understand. But the book (cf. 159 page of Hatcher's book, algebraic topology) explain the case where $m=2$. In this case $g_\ast$ is not injective.

I cannot understand the above statement. From the above assumption when $m=2$, I can infer that $X$ is $\vee_{i\in I} S^1$ so that $g_\ast$ is just a permutation of elements of $I$.

Hence $g_\ast$ is injective. Where is wrong in my argument ?

Thank you in advance.

HK Lee
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1 Answers1

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You've misread the question.

Of course the homomorphism $g_*$ is always injective and surjective: taking homology is functorial, and the action of $g$ is a homeomorphism, so $g_*$ is an isomorphism.

The question asks if the map $G \rightarrow GL_n(\mathbb{Z}/2)$ is injective, i.e. do we get a faithful representation. In other words: can we find two different group elements that give the same automorphism of $GL_n(\mathbb{Z}/2)$. (I assume also that the question asks for $G$ to already be a subgroup of homeomorphisms, otherwise the question is trivial as we can make $G$ act trivially.)

Now that you know the correct question, do you think you can do it?

Dylan Wilson
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    Thank you. I misreaded. Consider $X = { (x,y,z)\in S^2(1) | x=0$ or $y=0 }$. And assume that $X$ has six vertices which have pairwise equidistance. And $G={\bf Z}2$ so that the action is : $(0,0,1)\rightarrow (0,0,-1)$ and other points is not moved. Note that $X = \vee{i=1}^3 S^1$. And the representation is not faithful. – HK Lee Apr 11 '13 at 10:01