Exactness of completed tensor product of nuclear spaces Let  $0 \to V \to W \to L \to 0$ be a strict short exact sequence
of (complete) nuclear spaces, i.e. it is a short exact sequence of
(complete) nuclear spaces, all the maps are continuous,  the map  $V  
\to W$  is a closed embeding, the topology on $V$ is induced from
$W$ and the map $W \to L $ is open. Let $U$ be a (complete) nuclear
space.
Is it true that  the sequence obtained by completed tensor product with $U$
(i.e. $0 \to V \hat{\otimes}  U \to W \hat{\otimes}  U  \to L  
\hat{\otimes}  U  \to 0$)
is also strict short exact sequence?
We know that this is true if all the spaces are Frechet or if all
the spaces are dual Frechet, but is this true in general?
 A: Unfortunately, this is not true in general.
Let $E$, $F$, $G$ and $H$ be complete nuclear lcs such that
$$ 0 \longrightarrow E \stackrel{S}{\longrightarrow} F \stackrel{T}{\longrightarrow} G \longrightarrow 0 $$
is a strict short exact sequence. Note that the sequence
$$ 0 \longrightarrow E \mathbin{\hat\otimes} H \stackrel{S \mathbin{\hat\otimes} \text{id}_H}{\longrightarrow} F \mathbin{\hat\otimes} H \stackrel{T \mathbin{\hat\otimes} \text{id}_H}{\longrightarrow} G \mathbin{\hat\otimes} H \longrightarrow 0 $$
is a strict short exact sequence if and only if $T \mathbin{\hat\otimes} \text{id}_H : F \mathbin{\hat\otimes} H \to G \mathbin{\hat\otimes} H$ is surjective; everything else is automatic by the mapping properties of the injective resp. projective topology.
Let $w$ and $w*$ denote the weak and weak-$*$ topologies, and let $T' : G_{w*}' \to F_{w*}'$ denote the adjoint of $T$.
It follows from [Sch99, §IV.9.4] that $F \mathbin{\hat\otimes} H \cong \mathfrak{B}_e(F_{w*} \times H_{w*})$. The latter is linearly isomorphic with $\mathfrak{L}(F_{w*}' , H_w)$, and the following diagram commutes:
$$\require{AMScd}
\begin{CD}
F \mathbin{\hat\otimes} H @>T \mathbin{\hat\otimes} \text{id}_H >> G \mathbin{\hat\otimes} H \\
@VV V @VV V \\
\mathfrak{L}(F_{w*}' , H_w) @> R \mapsto RT' >> \mathfrak{L}(G_{w*}' , H_w) \\
\end{CD}$$
Since $G \cong F/E$, we have $G' = E^\perp$, and the weak-$*$ topology on $G'$ coincides with the relative $F_{w*}'$ topology of $E^\perp$. Thus, we see that $T \mathbin{\hat\otimes} \text{id}_H$ is surjective if and only if every weak-$*$-to-weak continuous operator $E^\perp \to H$ can be extended to $F'$. This is not always the case:

Counterexample. Let $E$ be a closed, non-complemented subspace of a nuclear Fréchet space $F$.¹ Furthermore, let $G := F/E$, and let $H := G_\beta'$ be the strong dual of $G$. Then $E$, $F$ and $G$ are nuclear Fréchet spaces (so in particular reflexive), and $H$ is a complete, nuclear (DF)-space.
In a Fréchet space, all weakly complemented subspaces are complemented, so $E$ is not weakly complemented in $F$. By duality, $E^\perp = G'$ is not weak-$*$ complemented in $F'$. Since $G$ is reflexive, the weak and weak-$*$ topologies on $H = G_\beta'$ coincide, so we have $\text{id}_H \in \mathfrak{L}(G_{w*}' , H_w)$. Since $G'$ is not weak-$*$ complemented, the map $\text{id}_H \in \mathfrak{L}(G_{w*}' , H_w)$ has no extension in $\mathfrak{L}(F_{w*}' , H_w)$.

Convesely, if $E$ is (weakly) complemented, then $T \mathbin{\hat\otimes} \text{id}_H$ is always surjective.
¹: Concrete examples of this kind are given in this answer, or in [MV97, Exercise 31.4], or in [DM76], among others.
References.
[DM76] Plamen Djakov, Boris Mitiagin, Modified construction of nuclear Fréchet spaces without basis, Journal of Functional Analysis, vol. 23 (1976), issue 4, pp. 415–433. DOI: 10.1016/0022-1236(76)90066-5.
[MV97] Reinhold Meise, Dietmar Vogt, Introduction to Functional Analysis (1997), Oxford Graduate Texts in Mathematics, Clarendon Press, Oxford.
[Pie72] Albrecht Pietsch, Nuclear Locally Convex Spaces (1972), Ergebnisse der Mathematik und ihrer Grenzgebiete, Springer, Berlin.
[Sch99] H.H. Schaefer, M.P. Wolff (translator), Topological Vector Spaces, Second Edition (1999), Springer Graduate Texts in Mathematics, Springer, New York.
