Below I describe a proof that the dimension of the row space of a matrix is equal to that of its column space. I would like to know if the proof is correct.
Let $M$ be an $m\times n$ matrix. Let $F$ be a field, and let $V=F^m$ and $W=F^n$ be the $m$ and $n$ dimensional $F$-coordinate spaces respectively. We want to show that $M^TV$ and $MW$ have the same dimension.
Strategy: Show that there is a surjective homomorphism from $M^TV$ onto $(MW)^*$, where $*$ refers to the dual, so that $\text{dim}(M^TV)\geq\text{dim}(MW)^*=\text{dim}(MW)$. By symmetry, we also get the reverse inequality, so $\text{dim}(M^TV)=\text{dim}(MW)$.
For $v\in V$ and $w\in W$, we have that $v^T(Mw)=(v^TM)w=(M^Tv)^Tw$.
Let $f:V\rightarrow (MW)^*$ be defined by $f(v)=(Mw\mapsto v^T(Mw))$
Then $f$ is a surjective homomorphism, because $MW\subseteq V$.
So we have that $f(v)(Mw)=(M^Tv)^Tw$ for all $v\in V, w\in W$.
Let $\phi:M^TV\mapsto (MW)^*$ be defined by $\phi(M^Tv)=f(v)$. We show that $\phi$ is well-defined.
Let $v_1,v_2\in V$. Suppose $M^Tv_1=M^Tv_2$. Then for all $w\in W$, $f(v_1)(Mw)=(M^Tv_1)^Tw=(M^Tv_2)^Tw=f(v_2)(Mw)$.
It follows that $\phi(M^Tv_1)=f(v_1)=f(v_2)=\phi(M^Tv_2)$.
Furthermore, $\phi$ is a homomorphism and it is surjective onto $f(V)=(MW)^*$, so $\phi$ is the desired surjective homomorphism and we are done. $\blacksquare$
Yes, this is all correct. There's not a lot to say about this beyond that, and since your uneasiness is vague, it's difficult to rectify any one point in the proof. There's just one point that I want to highlight:
This justification strikes me as being a little wide of the mark. The assertion that $f : V \to (MW)^*$ is a surjective homomorphism contains a few suppressed steps, the most important being $f$ is surjective and $f$ is a homomorphism, as well as some trivial ones like $f$ being well-defined, $f(v)$ being a well-defined map, and $f(v)$ being linear. None of these steps are important enough to explain individually in the proof, but your justification points to $f(v)$ being well-defined, which seems like one of the less important things to highlight.
It's not a flaw, really, but ideally you want your one line justifications to be incisive, pointing to the most tricky parts of what you're glossing over. It usually also indicates a lack of confidence in the proof, which may be why you're feeling generally uneasy.
The map $f$ is actually a composition of two well-known, useful maps in functional analysis. It is the composition of the duality map $J$ on $V$, which takes a vector $v$ in $V$ and maps it to $x \mapsto v^\top x$ in $V^*$ (this actually does have an analogue in more general normed linear spaces, but it is rarely linear and sometimes multivalued outside of Euclidean or Hilbert spaces), and the restriction map: $$f \in V^* \mapsto f|_{(MW)^*}.$$ Both of these maps are linear. The Riesz representation theorem states the duality map is surjective, and the Hahn-Banach extension theorem states that the restriction map is also surjective. So, the composition is linear and surjective.
It's just another way to see it. Maybe it'll help you feel more confident? Either way, as I said, the proof is good.