Request:  intermediate-level proof: every 2-homology class of a 4-manifold is generated by a surface. Hi, everyone:
For the sake of context, I am a graduate student, and I have taken classes in
algebraic topology and differential geometry. Still, the 2 proofs I have found
are a little too terse for me; they are both around 10 lines long, and each line
seems to pack around 10 pages of results. Of course, I am considering cases for
"reasonable" spaces, being the beginner I am at this point.
It would also be great if someone knew of similar results for H_1 (equiv. H_3).
Thanks in Advance.
 A: Torsten's answer was good, but there are also more elementary answers.  Here's one, which is essentially a big transversality argument, followed by a mild de-singularization.
Let me consider $M$ to be a triangulated 4-manifold, and represent a class in $H_2(M)$ as a sum of 2-faces of the triangulation.  For each 2-face appearing with multiplicity $n$ in the sum, take $n$ copies of the face, pushed off of each other slightly and meeting only at the edges.
Now, along each edge (1-face) of the triangulation, since we started with a 2-cycle the total number of triangles meeting there is $0$.  (This is a signed count if we are working in $H_2(M; \mathbb{Z})$, and means that there are an even number if we are in $H_2(M; \mathbb{Z}/2)$.)  Along each edge, pair up the incident triangles in an arbitrary way that's compatible with the orientations, and resolve the intersections along the interior of the edge according to that pairing.  (A neighborhood of the edge looks like $D^3 \times I$; the incident triangles are coming in from fixed directions, i.e., at fixed points in $S^2 \times I$; so given any pairing of the points on $S^2$, we can just join them up and avoid the edge altogether.  It's easier to think about what happens in a 3-manifold, where it's very similar, you just have to be more picky about how you pair the incident triangles.)
After the last step, we have a surface $S$ with codimension-2 singularities at points in $M$.  For each such singularity, consider a small ball $B$ in $M$ and consider $S \cap \partial B$.  This is a link $L$ in a 3-manifold (oriented or not, depending on which homology we look at).  Replace $S \cap B$ with a Seifert surface for $L$, and we're done.  (You could also use any surface with boundary $L$ inside the 4-ball $B$ instead of the Seifert surface, of course.  Frequently you can get lower genus that way.)
A: I do not know if this provides more details: 
I assume the closed ($4$-)manifold $M$ to
be oriented (we need the submanifold to be oriented to have an integral homology
class and the construction I am about to give will make the normal bundle
oriented) and as transversality arguments are a little bit tricky in the purely
topological case I shall also assume $M$ is smooth. If $b\in H_2(M,\mathbb Z)$
we let $c\in H^2(M,\mathbb Z)$ be its Poincaré dual. The exponential sequence of
sheaveas $0\rightarrow\mathbb Z\rightarrow\mathcal C_M\rightarrow\mathcal
C_M^*\rightarrow0$ of sheaves on $M$ (where $\mathcal C_M$ is the sheaf of
continuous complex-valued functions) and the fact that the cohomology of
$\mathcal C_M$ is trivial (because of existence of partitions of unity) shows
that $c$ may be represented as the first Chern class of a complex line bundle
$L$. A general section of this line bundle is transversal to the zero section
and hence its zero set is a $2$-submanifold of $M$. Its normal bundle is the
restriction of $L$ to $S$ and is therefore orientable and hence so is $S$. Its
class is then equal to $b$.
Essentially the same argument (but with a slightly different exponential
sequence) shows that every class $w$ in $H^1(M,\mathbb Z/2)$ is the
Stiefel-Whitney class of a real line bundle and the zero set of a general
section will have its class equal to the Poincaré dual of $w$ in $H_3(M,\mathbb
Z/2)$ and hence any such class can be represented by a non-orientable
$3$-dimensional submanifold. As for a class in $H_3(M,\mathbb Z)$ its Poincaré
dual in $H^1(M,\mathbb Z)$ is the inverse image of the fundamental class in
$H^1(S^1,\mathbb Z)$ of a map $M \rightarrow S^1$ (this is because every element
of $H^1(X,\mathbb Z)$ for a nice enough $X$ is represented in that way). The
inverse image of a regular value of this map (which exists by Sard's lemma) will
then be a $3$-dimensional orientable (in fact with trivial normal bundle)
submanifold of $M$ representing the given class in $H_3(M,\mathbb Z)$. Finally,
elements of $H_1(M,\mathbb Z)$ are easy to deal with. By the Hurewicz theorem
any class of $H_1(M,\mathbb Z)$ is the image of the fundamental class of $S^1$
under a map $S^1 \rightarrow M$. As the dimension ($1$) of $S^1$ is smaller than
half ($2=4/2$) the dimension of $M$ a general such function will be an embedding.
