Existence of a Surjective Linear Map $\mathbb{R}^3$ to $\mathbb{R}^4$? Is it possible for there to be a surjective linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^4$?

I feel like this isn't possible because $\mathbb{R}^3$ has fewer vectors than $\mathbb{R}^4$. If so, linear transformations between the two spaces should be injective. But if that's the case, for all linear transformations $\mathbb{R}^3\to\mathbb{R}^4$, the kernel should only ever contain the zero vector. But this is not the case. Why?
There are functions from $\Bbb R^3$ to $\Bbb R^4$ which are surjective, but none of them preserve the vector space structures. By the rank-nullity formula: $$3 = \dim \ker T + \dim {\rm Im } \ T,$$ so we can't have $\dim {\rm Im } \ T = 4 $.
They have the same cardinality but I don't believe cardinality implies existence of surjective linear transformation.