Endomorphisms: if $Im(g)$ is contained in $Im(f)$ then necessarily $g=f \phi$? Let $f$ and $g$ be endomorphisms of a vector space $V$. 
If $Im(g)$ is contained in $Im(f)$ then does there necessarily exist an endomorphism $\phi$ of $V$ such that  $g=f \phi$?
Do we also have that $Ker(g)$ contained in $Ker(f)$ implies the existence of an endomorphism $\phi$ such that $f=\phi g$?
 A: Let $\{e_i\mid i \in I\}$ be a basis for $V$.

Since $\text{im}(g) \subseteq \text{im}(f)$, it follows that for each $i \in I$, there exists $x_i \in V$ such that $f(x_i) = g(e_i)$.

Then let $\phi$ be the unique endomorphism of $V$ such that $\phi(e_i) = x_i$, for all $i \in I$.

It follows that $g = f \circ \phi$.
A: Yes, and the answer writes itself.
Chose a set of $g(x_i)$ which is a basis for $Im(g)$. The set of $x_i$ are linearly independent, and you can extend them to a basis of $V$.
Then pick an element of $y_i\in f^{-1}(g(x_i))$ and declare $\phi (x_i)=y_i$ for your original $x_i$'s and map the remaining basis elements however you like.
This determines a unique linear transformation with the property you asked for.
A: Hint : The hypothesis already tells you that for all $v\in V$, $g(v)\in Im(f)$, i.e. there exist $w\in V$ such that $f(w)=g(v)$. Use this to construct first a function $\phi$ on a basis of $V$, and then extend that function to a linear application defined on $V$.
For the second part, first take a basis $\mathcal{B}_1$ of $\ker g$, and then extend it to a basis $\mathcal{B}$ of $V$. Then the complement $\mathcal{B}_2=\mathcal{B}\setminus \mathcal{B}_1$ is a linearly independent family, and $g(\mathcal{B}_2 )$ is a basis of the image of $g$. Extend it to a basis $\mathcal{B}'$ of $V$, and then you can simply define $\phi $ by putting $\phi(g(b))=f(b)$ for all $b\in \mathcal{B}_2$ and anything you want for the other vectors in your basis $\mathcal{B}'$. You can now check that $f=\phi\circ g$ by comparing the values on $\mathcal{B}$ : on $\mathcal{B}_1$ both are zero by hypothesis, on $\mathcal{B}_2$ they are equal by construction.
