Consider a linear transformation $T\colon V\to W$, for vector spaces $V$ and $W$. The problem is to prove, without using the rank-nullity theorem,
- If $T$ is injective, then $\dim{V}\leq\dim{W}$,
- If $T$ is surjective, then $\dim{V}\geq\dim{W}$.
I argued as follows.
Let $m=\dim{V}$ and $n=\dim{W}$. Suppose $T$ is injective, and suppose to the contrary that $m>n$. Every vector in $V$ can be as a linear combination of $m$ basis vectors and every vector in $W$ can be expressed as a linear combination of $n$ basis vectors.
If we choose an ordered basis of $V$ and $W$, and let every linear combination of the ordered basis of $W$ have its coefficients coincide with the first $n$ coefficients in the linear combination of the ordered basis of $V$, and let the remaining $m-n$ coefficients be $0$, we have a bijection between the vectors in $W$ and some of the vectors in $V$. But since there are obviously vectors in $V$ outside of this bijection, we must have $|V|>|W|$.
But now $T$ cannot be injective, contradiction. $\square$
I used a similar argument for the case where $T$ is surjective. My question is: is my deduction that $|V|>|W|$ by forming a bijection between $W$ and a proper subset of $V$ above valid? Is this a valid way to deal with infinite cardinalities? If not, how else can we solve the problem?