TITLE: Are nuclear operators closed under extensions?
QUESTION [6 upvotes]: Given $X_i, Y_i$ Banach spaces, $f_j, g_j, T_i$ bounded linear operators for $i=1,2,3$ and $j=1,2$. We have the following diagram
$\require{AMScd}$
\begin{CD}
0 @>>> X_1 @>f_1>> X_2 @>f_2>> X_3 @>>> 0\\
@V VV @V T_1 VV  @V T_2 VV @V T_3 VV @V VV \\
0 @>>> Y_1 @>>g_1> Y_2 @>>g_2> Y_3 @>>> 0
\end{CD}
with two horizontal topologically short exact sequences. If $T_1$ and $T_3$ are nuclear operators, does it imply that $T_2$ is nuclear as well? A reference to problems of this general form, would be most welcome.

REPLY [7 votes]: The answer is no: you can even have $T_1=T_3=0$ and $T_2$ equal to the identity $id$ on an infinite dimensional Banach space. 
Indeed, consider the following commutative diagram with exact rows:
$$\begin{CD}
0@>>> 0 @>0>>  X @>id>> X @>>> 0\\
&&@V0VV @VV{id}V @VV0V\\
0@>>>X  @>>id> X @>>0> 0 @>>> 0
\end{CD}
$$
See this paper for related results.

REPLY [5 votes]: Of course, Yemon was faster than me. But I want to emphasize that
the point is very elementary linear algebra: The simple commutative diagram
$\begin{CD}
0 @>>> \mathbb R @>f_1>> \mathbb R^2 @>f_2>> \mathbb R @>>> 0\\
@V VV @V T_1 VV  @V T_2 VV @V T_3 VV @V VV \\
0 @>>> \mathbb R @>>f_1> \mathbb R^2 @>>f_2> \mathbb R @>>> 0
\end{CD}$
with the natural inclusion $f_1(x)=(x,0)$ and projection $f_2(x,y)=y$ in the rows shows shows that $T_2$ is not at all determined by $T_1$ and $T_3$. If $T_2$ is given by a matrix $\begin{bmatrix} a&b\\ 0&c\end{bmatrix}$ you get nothing for $b$.