Representing second homology by embedded spheres Let $M$ be a simply connected 4-manifold (non-compact) with second homology group:
\begin{align*}
&H_2(X;\mathbb{Z})\cong\mathbb{Z}^k, 
\end{align*}
for some $k\geq1$.
Assume that every generator of $H_2$ can be represented by an embedded 2-sphere and the same holds for the sum of distinct generators (i.e. the classes of the form $(a_1,...,a_k)$ and $a_k=0,1$).
Is it true that any embedded two-sphere must represent one of the aforementioned classes?
EDIT: Thank you all for the useful comments.
Actually, I am just interested in the crepant resolution of $\mathbb{C}^2/\mathbb{Z}_k$ for all $k\geq2$ (the case k=2 should just be $T^{\ast}S^2$). Does anyone know if that is true in this setting?
 A: No, there are counterexamples, let $X= S^2 \times (D^2 \cup_\partial D^2)$. Here we think of $D^2$ as the unit ball. Let $f:S^2 \rightarrow S^2$ be a degree two map which we can think of as a map from $D^2 \rightarrow S^2$ given by the identification $B(O,1/2)/\partial B(O,1/2) \cong S^2$ and $D^2 - B(O,1/2)/\sim \:   \cong S^2$ where I identify the components of the boundary to two separate points. Here $B(O,r)$ is a ball of radius $r$ around the origin.
We use a $'$ to denote a coordinate in the second copy of $D^2$. Define a map $D^2 \rightarrow S^2 \times (D^2 \cup_\partial D^2)$ by $(x,y) \rightarrow (f(x,y),x,y)$. This is clearly injective. Now we can extend this map to a map $D^2 \cup_\partial D^2$ by sending the new copy of $D^2$ into $S^2 \times (D^2 \cup_\partial D^2)$ by the map $(x',y') \rightarrow (*,x',y')$ where $*$ is the point in $S^2$ that $f$ sends $\partial(D^2)$. This map is also injective. Since the two maps only share a codomain where we glue them together, they must be globally injective
This map $D^2 \cup_\partial D^2 \rightarrow S^2 \rightarrow S^2 \times (D^2 \cup_\partial D^2) $ is then an injective map which represents 2 times the canonical embedding $S^2 \rightarrow S^2 \times (D^2 \cup_\partial D^2)$ plus the inclusion into the other factor. You can see this by projecting to either factor and using the Kunneth formula.
You can make this example noncompact by deleting any point $\{x\} \times \{y\}$ where $x \neq *$ and $y$ is in the second copy of $D^2$. The the manifold is homotopy equivalent to a wedge of 2 2-spheres, and the map represents twice the inclusion of one of these two spheres plus the inclusion of the other, just as it did before.
This example is just a higher dimensional version of the (2,1) torus example.
A: Not a complete answer, but too long for a comment.
It might be true, but looking at the analogous situation in 2-dimensions makes me think not, for your hypotheses work for the torus $S^1 \times S^1$, but each $(p,q)$ torus knot (where $p$ and $q$ are relatively prime) is an embedded circle without being a $0/1$-combination of generators.
