Are $\operatorname{Ker}(T)$ and $W/\operatorname{Im}(T)$ isomorphic? Given a linear mapping $T:V\rightarrow W$ between the two vector spaces, $\operatorname{Im}(T)$ and $V/\operatorname{ker}(T)$ are isomorphic. I wonder if $\operatorname{Ker}(T)$ and $W/\operatorname{Im}(T)$ are also isomorphic?
If yes, then we also have $\dim{(W)}=\dim{(\operatorname{Ker}(T))} + \dim{(\operatorname{Im}(T))}$? 
Thanks!
 A: Another way to see that this cannot be true is to use the correct rank-nullity theorem, i.e., $\dim{(V)} = \dim{(\operatorname{Ker}{(T)})} + \dim{(\operatorname{Im}{(T)})}$. If we assume $\dim{(W)}=\dim{(\operatorname{Ker}{(T)})} + \dim{(\operatorname{Im}{(T)})}$, then we must have $\dim{(W)} = \dim{(V)}$!
Now, if we assume that $V$ and $W$ are vector spaces of the same dimension, and $T:V\to W$ is a linear map, then $V/\operatorname{Ker}{(T)}$ and $\operatorname{Im}{(T)}$ are canonically isomorphic via $[v] \to T(v)$, where $[v]$ is the class of $v\in V$ in $V/\operatorname{Ker}{(T)}$. 
Since $$\dim{(W)} = \dim{(\operatorname{Im}{(T)})} + \dim{(W/\operatorname{Im}{(T)})}$$
and by rank-nullity
$$\dim{(W)}=\dim{(V)} = \dim{(\operatorname{Ker}{(T)})} + \dim{(\operatorname{Im}{(T)})},$$
we may subtract both equations and obtain $\dim{(W/\operatorname{Im}{(T)})}=\dim{(\operatorname{Ker}{(T)})}$. Hence, $W/\operatorname{Im}{(T)}$ and $\operatorname{Ker}{(T)}$ are vector spaces of the same dimension, therefore isomorphic... but the isomorphism is not canonical! They are isomorphic simply because their dimensions coincide.
Although the term "canonical" does not have a very precise meaning in general, in this entry by canonical I mean "natural", or "coordinate-free". For instance, if $T:V\to W$ is any linear map of $\mathbb{F}$-vector spaces, then the isomorphism $\phi:V/\operatorname{Ker}{(T)}\to \operatorname{Im}{(T)}$ is canonical because I can describe it without having to choose bases for the vector spaces involved. The map is simply $\phi([v])= T(v)$, for any $v\in V$. However, as we have shown above, if $V$ and $W$ have the same dimension, we also have an isomorphism $\psi: W/\operatorname{Im}{(T)} \to \operatorname{Ker}{(T)}$, but we cannot describe this isomorphism unless I define a basis $\{w_1,\ldots,w_n\}$ of $W/\operatorname{Im}{(T)}$ and a basis $\{v_1,\ldots, v_n\}$ of $\operatorname{Ker}{(T)}$, and then, once in coordinates, we choose an arbitrary isomorphism $\mathbb{F}^n \to \mathbb{F}^n$.
