Intuition behind the injectivity part of Hurewicz Theorem The surjectivity part of Hurewicz Theorem is easy to understand: under the inductive hypothesis that all homotopy groups (of a CW-complex) up to dimension $n$ are trivial, it is clear (I believe) how to decompose any cycle of dimension $n+1$ into a sum of images of oriented spheres (of dimension $n+1$).
However, when I try to read some proof of the theorem, they start talking about commutative diagrams and exact sequences.  I fail to see the geometric intuition behind the injectivity part.  It is not even clear to me why an acyclic simply-connected 2-dimensional finite CW-comples is contractible.  What is the intuition?

Update.
After thinking more about it, it seems not that hard, and it can be viewed as a straightforward generalisation of the 1-dimensional case (that $H_1(X)$ is isomorphic to the abelianisation of $\pi_1(X, x_0)$).
The key idea is to construct an inverse homomorphism "by hand," starting from a suitable map $\rho$ defined on the set of oriented $(n+1)$-cells of the given CW-complex $X$, which to every oriented $(n+1)$-cell would associate an element of $\pi_{n+1}(X, x_0)$, and which would associate inverse elements of $\pi_{n+1}(X, x_0)$ to opposite orientations of the same $(n+1)$-cell.
To construct $\rho$, start by fixing a homotopy of the identity map of $X$ to a map that sends the $n$-skeleton of $X$ to a point/vertex. (Such a homotopy exists because the homotopy groups of $X$ are trivial up to dimension $n$, and because of the homotopy extension property.)
Such a map $\rho$ obviously induces a homomorphism
$$
\sigma\colon C_{n+1}(X)\to\pi_{n+1}(X, x_0)/[\pi_{n+1}(X, x_0),\pi_{n+1}(X, x_0)],
$$
where $C_{n+1}(X)$ is the Abelian group of $(n+1)$-dimensional cellular chains in $X$. (I know that $\pi_n(X,x_0)$ are Abelian for $n\ge 2$, but I do not want to exclude $\pi_1(X,x_0)$ here.) Moreover, if appropriately defined, it will be zero on the subgroup of $(n+1)$-dimensional boundaries. Thus, the homomorphism $\sigma$ will induce a homomorphism
$$
s\colon H_{n+1}(X)\to\pi_{n+1}(X, x_0)/[\pi_{n+1}(X, x_0),\pi_{n+1}(X, x_0)].
$$
Let
$$
p\colon\pi_{n+1}(X, x_0)/[\pi_{n+1}(X, x_0),\pi_{n+1}(X, x_0)]\to H_{n+1}(X)
$$
be the "extension" of Hurewicz homomorphism (if "extension" is not the right term, what is the right one?).
It is more or less clear that $s$ will be a right inverse of $p$ (assuming the right definition of $\rho$, which I did not give), which implies the surjectivity of Hurewicz homomorphism.
It looks like it can also be checked "geometrically" that $s$ will be a left inverse of $p$, with implies the injectivity of Hurewicz homomorphism.
(Incidentally, if $p\circ s$ is known to be identity, then to check that $s\circ p$ is an identity too, it is enough to check that the kernel of $s\circ p$ is trivial.)
I'd need to check details and to see if this argument can be presented better. I am not yet absolutely sure about it.
 A: The ''geometric intuition' part is to see that homology is an ''algebra'' but not ''topology''. The real topology there is done and make rigorous using algebra only.
In (simplicial/singular/cellular) homology, one is counting number of ''spherical objects'', where spheres are building blocks and simplest in terms of topology and algebra ($H_k(S^n)=(\delta_{kn}+\delta_{0k})Z$), then one proceed to really count with long exact sequence/Mayer Vietoris sequence. In homotopy group, one is counting number of ''spherical maps'', hence objects like Hopf's fibration is included and one again counts using long exact sequences.
There is a result showing that if $X$ is n-connected, it can have a cellular decomposition of no cells of dimension below $n+1$. Therefore, the long exact sequence for working in homology or homotopy simply coincide for degree below $n+1$ where spheres remain ''simple'' in both cases. By teh five lemma, they simply coincide, and hence the Hurewicz theorem.
With this, to explain why acyclic complex is contractible, this is really trying to understand why weakly homotopy equivalent space are homotopy equivalent.
