How can q(z,l,u,c|x) be factorized to q(z|x)q(l|x)q(c|z)q(u|c,z)? How can q(z,l,u,c|x) be factorized to q(z|x)q(l|x)q(c|z)q(u|c,z)?
I don't know what assumptions are behind this equation, which is I want to know. I think normal chain rule of probability is in the process, but only chain rule can't derive the formula.
What I know is that z, l, u follows gaussian distribution, and c follows multinomial distribution.
It's from the paper Xu C, Lopez R, Mehlman E, Regier J, Jordan MI, Yosef N. Probabilistic harmonization and annotation of single-cell transcriptomics data with deep generative models. Mol Syst Biol. 2021 Jan;17(1):e9620. doi: 10.15252/msb.20209620. PMID: 33491336; PMCID: PMC7829634.
That formula is used for derive ELBO, since the model is similar to VAE.
 A: It can't be factorised that way.  The paper doesn't claim it can. What they say is that they are computing a variational approximation using a product of that form. It's fairly common for variational approximations to be products, even when the true posterior or likelihood doesn't factorise as a product of independent components.
A: In general $P(XY | A) = P(X | YA) P(Y | A)$.
If $X$ and $Y$ are conditionally independent given $A$ then $P(X | YA) = P(X | A) $ and $P(XY | A) = P(X | A) P(Y | A)$.
For your problem, if you assume the appropriate conditional independence structure for your variables, you can write that factorization.
I haven't looked at that paper, but something that's extremely common is to make some assumptions about what's conditionally independent of what to make the model more tractable and useful. In the best case, the assumptions are true (or in some sense close to true)! In other settings, the assumptions are even clearly false (e.g. Naive Bayes applied to language) but still produces useful enough results compared to alternatives that the assumptions are deemed worth making.
I often come back to the aphorism of George P. Box that "all models are wrong, but some are useful."
A: It is a basic theorem of conditional probability that $$P(A,B \mid C) = P(A \mid B,C) \cdot P(B \mid C).$$
Furthermore, if $A$ and $B$ are conditionally independent given $C$, then $P(A \mid B,C) = P(A \mid C)$, and so $$P(A,B \mid C) = P(A \mid C) \cdot P(B \mid C).$$
Conversely, if $A$ depends on $C$ only via $B$, then $P(A \mid B,C) = P(A \mid B)$, and so $$P(A,B \mid C) = P(A \mid B) \cdot P(B \mid C).$$
The factorization $q(z,l,u,c \mid x) = q(z \mid x)q(l \mid x)q(c \mid z)q(u \mid c,z)$ given in your question can thus be interpreted as a description of how the various variables are assumed to (approximately) depend on each other.  In particular, we can see that:

*

*$z$ and $l$ are assumed to be conditionally independent, given $x$; and

*$c$ and $u$ are assumed to depend on $x$ only via $z$ (and to be conditionally independent of $l$).

(Note that we cannot say anything about the dependence between $c$ and $u$, except that it is assumed to be potentially non-negligible.  In particular, $q(c \mid z) q(u \mid c,z)$ and $q(c \mid u,z) q(u \mid z)$ are both equally valid factorizations of $q(c,u \mid z)$.  Remember that statistical dependence is not causality, and always goes both ways, even when the notation might make it look otherwise.)
