Using a Quotient map to induce a well-defined map Let $X, Y, Z$ be Banach spaces, and let $q:X\to Y$ be a quotient map.
Given a bounded linear operator $T:X\to Z$, does there exist some criteria on $T$ for determining when $q$ can be used to induce a well-defined $\hat{T}:Z\to Y$?

Definition Let $X,Y$ be Banach spaces.  A bounded linear operator $q:X\to Y$ is a quotient map if
1) $q$ is surjective and
2) For all $y\in Y$, $\|y\| = \text{inf}\{\|x\|:q(x) = y\}$

So under the hypothesis of my question, $Y$ is a quotient of $X$ in the sense that there is a quotient map from $X$ to $Y$.
 A: $T$ has to be surjective, $z=T(x)=T(x')$ we must have $q(x)=q(x')$. Thus $T(x)=0$ implies $q(x)=0$.


*

*$T$ surjective

*$Ker T\subset Ker q$.
We set $q'(z)=q(x)$ where $T(x)=z$.
A: When $q: X\rightarrow Y$ is bounded linear, $T:X\rightarrow Z$ is bd linear and surjective and $\ker T \subset \ker q$ this suffices to provide a unique bounded linear operator $\hat{T}: Z\rightarrow Y$ with $q=\hat{T} \circ T$. (Sorry, I repeat here most of what Tsemo said).
To define $y=\hat{T}(z)$ you pick $x$ so that $T(x)=z$ and set $y=q(x)$. This is independent of the choice of $x$ because of the kernel condition. It is also straigtforward to see that the obtained map is linear. 
To see that it is bounded you need to know the open mapping theorem: Since $T$ is a surjective Banach space mapping, it is open whence there is $r>0$ so that $TB_X(0,r)$ contains $B_Z(0,1)$. For a given $z$ you may then pick $x$ of norm not greater than $r\|z\|$ and then $\|\hat{T}z\| =\|q(x)\|\leq \|q\| r\|z\|$.
$q$ being surjective implies that also $\hat{T}$ is surjective.
