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Im trying to solve a Problem from Hatcher: Compute the simplicial homology groups of the $\Delta$-complex obtained from n+1 simplices $\Delta_0^2,\Delta_1^2,...,\Delta_n^2$ by identifying all three edges of $\Delta_0^2$ to a single edge ($=:e_0$) , and for $i>0$ identifying the edges $[v_0,v_1]$ and $[v_1,v_2]$ of $\Delta_i^2$ to a single edge ($=:e_i$) and the edge $[v_0,v_2]$ to the edge $[v_0,v_1]$ of $\Delta_{i-1}^2$.

I can calculate that $\partial_2 \Delta_0^2 = e_0$ and for $i>0\ $ $\partial_2 \Delta_i^2 = 2e_i-e_{i-1}$ and $\partial_1 e_i =0$ which yields:

$H_1(X) \cong \frac{ker\partial_1}{Im\partial_2}\cong \frac{ \mathbb{Z}<e_0,e_1,...,e_n>}{<e_0,2e_1-e_0,2e_2-e_1,...,2e_n-e_{n-1}>}$

and i cant simplify this. The correct solution is supposedly $\mathbb{Z}_{2^n}$.

I tried to tinker around with lower dimensional solutions and while i understand $\frac{\mathbb{Z}<e_0,e_1>}{<e_0,2e_1-e_0>} \cong \mathbb{Z}_2$ by rewriting the denominator as $<e_0,2e_1-e_0>=\mathbb{Z}(1,0)\oplus\mathbb{Z}(-1,2)\cong \mathbb{Z}(1,0)\oplus\mathbb{Z}(0,2)= <e_0,2e_1>$, i cant make sence of it with three generators: $\frac{\mathbb{Z}<e_0,e_1,e_2>}{<e_0,2e_1-e_0,2e_2-e_1>}$.

Could somebody please explain the case of three generators?

I would be very thankful for any help!

NoIdea
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    How's your linear algebra? Do you know how to reduce matrices to smith normal form (over $\mathbb Z$)? – Lee Mosher Dec 23 '23 at 21:51
  • My linear algebra should be fine. But i kind of skipped over $\textbf{Z}$-Moduls when everything was online during the pandemic. Smith normal form is essentially just singular value decomposition, correct? Do you mean i should just directly prove the claim that it is equal to the supposed solution? It would feel a bit cheesy because usually i dont already know the solution. Thats why i tried to see it with change of basis. – NoIdea Dec 23 '23 at 22:15
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    That's exactly what Smith normal form does, although it does so in a directed manner: it let's you compute a change of basis so that the quotient is obvious. Row operations do not change the row span; column operations change the basis but not the isomorphism class of the row span. – Lee Mosher Dec 23 '23 at 22:18
  • I managed to solve it using Smith normal form. Thanks very much for this hint! Smith normal form wasnt something i had on the radar. – NoIdea Dec 23 '23 at 23:20
  • This question has been asked and answered before! – FShrike Dec 23 '23 at 23:55

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