On well defined map when sending vector to its coordinate vector Let $T:V \to V$ be a linear operator on an $n$ dimensional vector space $V$ with a basis $B$. Set $A = [T]_B$ the representation matrix. Show that the map $$f:\text{ker}(T - \lambda I_V) \to \text{null}(A - \lambda I_n)$$ where one sends $\mathbf{w}$  domain to the coordinate vector $[\mathbf{w}]_B$ in the codomain. $$\mathbf{w} \mapsto [\mathbf{w}]_B$$
Show this map is an isomorphism.
Now this problem is not difficult. However one part puzzles me (a bit). As usual my first step usually is to show well defined ness of the map using the standard way : If $w_1 = w_2$, then show $f(w_1) = f(w_2)$.
I would naively say that since $f(w_1) = [w_1]_B = [w_2]_B = f(w_2)$ then we are done. However my professor used another way and more complicated in this question, so i think my way might be flawed. He showed that for every element in the domain , it also belongs to the codomain.
So does my way really show me that given an element $w $ in the domain, the image $[w]_B$ really does belong to the codomain? Or is my method wrong, please advice!
 A: See my comment in this answer about what you need to do to check that a function/map is well-defined.  In particular, when you have a function $f: X \to Y$ defined by "for all $x\in X$, $f(x) = \text{[some expression depending only on $x$]}$", you only need to check that $f(x) \in Y$.

As usual my first step usually is to show well defined ness of the map using the standard way : If $w_1 = w_2$, then show $f(w_1) = f(w_2)$.

That statment is true for every function $f$. It's true by substitution. It doesn't show well-definedness, or any other property.

I would naively say that since $f(w_1) = [w_1]_B = [w_2]_B = f(w_2)$ then we are done.

As noted above, you never have to do this step when the parameters to your function are simple variables. If you defined your function differently, like "For all $w\in W, v\in V$, let $f(w+v)=\dots$" then the parameter to $f$ isn't a simple variable. You'd have to check whether $w+v$ gives the whole domain, and that you get the same answer when $w_1+v_1=w_2+v_2$, because when you plug that value ($w_1+v_1$ or $w_2+v_2$) into $f$, you'd better get the same answer coming out.
See the link above for some examples of when you have to check this stuff.
So, in your case, we only have to check that when you put a vector into $f$, you get a vector in the range/codomain, i.e. check that for all $\mathbf w\in \text{ker}(T - \lambda I_V)$ we have $f(\mathbf w) \in \text{null}(A - \lambda I_n)$.  This is the work that your professor did.
