For Banach spaces $E$ and $F$ we denote the approximate operators by $\mathcal A(E,F)$ and projective tensor product by $\hat\otimes$. 


Consider the natural map 
$$\Delta: \mathcal A(\ell^q,\ell^p)\hat\otimes\mathcal A(\ell^p,\ell^q)\rightarrow \mathcal A(\ell^p), \quad S\otimes T\mapsto ST$$ 


Can we write elements in image of $\Delta$ in form of $TS$ where $T\in\mathcal A(\ell^q,\ell^p)$ and $S\in\mathcal A(\ell^p,\ell^q)$?



About representations of some elements in $\mathcal A(\ell^p)$
So I think this works. Fix $1<r<\infty$ and let $r'$ be the conjugate index to $r$ so $1/r + 1/r' = 1$.


Let
$$ \tau = \sum\_n S\_n \otimes T\_n \in \mathcal{A}(\ell^q,\ell^p) \widehat\otimes \mathcal{A}(\ell^p, \ell^q), $$
where by definition, $\sum \|S\_n\| \|T\_n\| < \infty$. By rescaling the $S\_n$ and $T\_n$ we may suppose that
$$ \sum\_n \|S\_n\|^{r'} < \infty, \quad \sum\_n \|T\_n\|^r < \infty. $$


Define $T:\ell^p \rightarrow \ell^r(\ell^q)$ by $T(x) = \big( T\_n(x) \big)$. This makes sense as
$$ \|T(x)\|^r = \sum\_n \|T\_n(x)\|^r \leq \|x\|^r \sum\_n \|T\_n\|^r
\qquad (x\in \ell^p). $$
For later, note that if $N$ is large and $T'(x) = (T\_1(x), T\_2(x), \cdots, T\_N(x), 0, 0, \cdots)$ then $\|T(x) - T'(x)\| \leq \|x\| \Big(\sum\_{n>N} \|T\_n\|^r\Big)^{1/r}$ and so $T'$ approximates $T$ *in norm* if $N$ is large.


Similarly we have $S:\ell^{p'} \rightarrow \ell^{r'}(\ell^{q'})$ given by $S(y) = \big( S\_n^\*(y) \big)$, using that $S\_n^\*: \ell^{p'} \rightarrow \ell^{q'}$ and that $\|S\_n^\*\| = \|S\_n\|$ for each $n$. Then a simple calculation shows that
$$ S^\* : \ell^r(\ell^q) \rightarrow \ell^p; \quad (x\_n) \mapsto \sum\_n S\_n(x\_n). $$
Thus $S^\*T= \sum\_n S\_n T\_n = \Delta(\tau)$.


Let $r=q$ and use the natural isomorphism $\ell^q(\ell^q) \cong \ell^q$ to see that $T:\ell^p\rightarrow \ell^q$ and $S^\*:\ell^q\rightarrow\ell^p$. As the isomorphism is compatible with the dual pairing, we still have $S^\*T=\Delta(\tau)$.


It remains to argue that $T,S^\*$ are approximable (limit of finite-rank) operators. It is clear that $T'$ (as defined above) is approximable, and so as $\mathcal A(E,F)$ is *closed* for any Banach spaces $E,F$, we see that $T$ is indeed approximable. Thus so is $S$, and hence $S^\*$ by a well-known theorem (I don't recall who to attribute this to, but it follows from the Principle of Local Reflexivity).


