I have heard lately that there is no smooth map from $S^2$ to any manifold that has constant rank $1$.
The proof for that claim is supposed to be based on the fact that the tangent bundle of $S^2$ is irreducible.
However, I do not know what irreducible means in this context and how the implication follows (I searched).
Any help?
A line bundle $E$ over a simply-connected manifold $N$ must be trivial. One way to see this is to choose a metric on $E$ and a metric connection $\nabla$. The connection must be flat (because $E$ is one-dimensional) and has no monodromy because $N$ is simply connected so the parallel transport maps $P_{p,q} \colon E_p \rightarrow E_q$ (where you can choose any path connecting $p$ and $q$) give us an isomorphism between $N \times E_p$ and $E$.
In your case, if $f \colon S^2 \rightarrow M$ is a smooth map of constant rank one, the kernel $\ker(df)$ must be a trivial line subbundle of $TS^2$ so you have a non-vanishing vector field on $S^2$, a contradiction.