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?
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 $\gamma$.