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I am concerned with 1-dimentional cycles in an n-manifold but the question can be generalised.

When I first studyed homology I thought that a manifold with homology groups $H_1=0$ (or in genral $H_i=0$ for i>0) must have no generators for the foundamental group. This is obviously wrong and a counterexample is the Poincaré homology sphere.

The reason I was convinced of the above statement is because:

1-$Z_1=im\partial_1$ , is the set of all cycles, including non path connected union of cycles. If A is a generator for the fondamental group, since loop are cycles then A is a cycle and must be in $Z_1$.

2- each element of $B_1=ker\partial_2$ (boundaties) can be seen as union of cycles that contain a simply connected surface, belonging to the manifold, for which they are the boundaries. Each of this cycles is therefore null homotopic although each of them may be homotopic to a different point (constant loop). If A is a generator for the fondamental group then A must not be in $B_1$ because it is not null homotopic.

3- To me, if $H_1=0$ then each element of $Z_1$ must be also in $B_1$ (they are exacly the same linear spaces) and therefore each cycle (i.e.each loop) must be null homotopic. Or, wich is the same, if A is a generator for the fondamental group then A must be in $Z_1$ but not in $B_1$ an this should imply $H_1\ne 0$.

My point is that, althought infinite, with a Poincaré homology sphere which has a foundamental group of 120 elements, $Z_1$ should be somehow 120 times bigger then $B_1$, which is not the case.

Where my reasoning is wrong? Can somebody help me to understant how it works and possibly give me the geometrical intuition of it?

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    $\begingroup$ The relationship between $\pi_1$ and $H_1$ for path-connected spaces is given by Hurewicz's theorem: $H_1$ is the abelianization of $\pi_1$. The construction you outlined gives a natural transformation $\pi_1 \to H_1$, but note that while the formal sum of cycles is commutative, composition of loops is not. So $H_1$ is abelian, but $\pi_1$ need not be, and the Hurewicz theorem says that this is essentially the only difference. $\endgroup$ – JHF Aug 24 '17 at 17:57
  • $\begingroup$ I would suggest you try to actually write down the details of what you are doing in step 2. In particular, how are you getting from a loop which is a sum of nullhomotopic loops when considered as a cycle to a nullhomotopy of the loop itself? $\endgroup$ – Eric Wofsey Aug 24 '17 at 19:05
  • $\begingroup$ @JHF. Thanks for your answer. I understand your algebraic point but I still miss the geometric intuition. To have a non trivial fundamental group I need at least one non null homotopic loop A. if A is a loop then it is in $Z_1$. if $H_1=0$ then each element of $Z_1$ is also in $B_1$. if A is in $B_1$ then it must be null homotopic and cannot be a generator. Where my reasoning is wrong? Maybe the answer is in the proof of the Hurewicz's theorem but I am not familiar with it. I will try to have a look at it. $\endgroup$ – vinardo Aug 24 '17 at 21:21
  • $\begingroup$ @EricWofsey. In step 2 I am just saying that if a loop is null homotopic then it must be a cycle in $B_1$. This is all I need and maybe I used too many words for it and made it confusing. It is then true that in $B_1$ there are cycles that are union of null homotopic separate loops. But this is not important to make my point. $\endgroup$ – vinardo Aug 24 '17 at 21:37
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If $\gamma$ is a loop whose representative as a cycle in $Z_1$ is a boundary, $\gamma$ is not necessarily nullhomotopic.

I think it's instructive to work through a simple example. Let's consider $S^1 \vee S^1$ be the figure eight, where the two simple loops are labelled $a$ and $b$, and study the composite loop $\gamma = aba^{-1}b^{-1}$. (This doesn't satisfy all your criteria: I don't know of an easily visualized space with perfect fundamental group.) This loop is not nullhomotopic, but it is mapped to the zero cycle $a + b - a - b = 0$, which is clearly a boundary. But there's no way to use the fact that $a + b - a - b$ is a boundary to construct a disk in $X$ that bounds $\gamma$.

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    $\begingroup$ To get that $\gamma$ is nullhomotopic, you need it to not just bound a surface but a disk. It actually does bound a surface, namely a torus with a disk removed. In fact, every nullhomologous loop bounds a surface. $\endgroup$ – Eric Wofsey Aug 24 '17 at 22:12
  • $\begingroup$ @JHF. I agree, Point 2 in my reasoning must be the one wrong. However your example is unsatisfactory because the figure 8 is not a manifold and most of all is 1-dimensional. In my reasoning at point 2 I was using (I should have been more clear) that the cycle is a boundary of simplexes or cells and therefore I need n>1. Can you really convince me that if $\gamma$ is a loop on an n-manifold whose representative as a cycle in $Z_1$ is a boundary (i.e. it is in $B_1$) then $\gamma$ is not necessarily nullhomotopic? I do not need to say n>1 here because I know there are not examples for n<3. $\endgroup$ – vinardo Aug 25 '17 at 0:12
  • $\begingroup$ @EricWofsey - Agreed. What I meant is, if a cycle is in $B_1$ then it bounds a disk although from the discussion above this seems not to be true. But I need somebody to convince me. $\endgroup$ – vinardo Aug 25 '17 at 0:24
  • $\begingroup$ @EricWofsey Corrected, thanks! $\endgroup$ – JHF Aug 25 '17 at 15:05
  • $\begingroup$ The "manifold" part is not so relevant here. You can consider instead the open subset of $\mathbb{R}^2$ with two points removed. That is a manifold, and it's homotopy equivalent to $S^1 \vee S^1$. I don't understand the rest of your comment. The cycle $a + b - a - b$ is a boundary of a $2$-chain, namely the empty chain. $\endgroup$ – JHF Aug 25 '17 at 15:14

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