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.



Are nuclear operators closed under extensions?
**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](http://www.eweb.unex.es/eweb/extracta/Vol-33-2/33J2Castillo.pdf) for related results. 


