Let $L=\mathbb{P}^l\subset\mathbb{P} ^ N \_ {\mathbb{C}}$ be a linear space and let $M=\mathbb{P}^{N-l-1}$ be a linear space skew to $L$, i.e. $L\cap M=\emptyset$. 
Let $X\subseteq\mathbb{P}^N\_{\mathbb{C}}$ be a closed irreducible variety not contained in $L$ and 
let $$ \pi\_L:X\dashrightarrow\mathbb{P}^{N-l-1}=M $$ be the linear projection, i.e. 
the rational map defined on $X\setminus L$ by $$ \pi\_L(x)=\langle L,x\rangle\cap M.$$
Let $x\in X\setminus L$ a point.


Is true that if $\overline{\pi\_L(X)}$ and $\overline{\pi\_L^{-1}(\pi\_L(x))}$ are smooth varieties, then $X$ is smooth at $x$?


Thanks.



A question on smoothness of varieties
The necessary condition is indeed the one given by Charles.


This is true if 
$$\dim X=\dim\overline{\pi\_L(X)}+\dim\overline{\pi\_L^{-1}(\pi\_L(x))}\tag{$\star$}$$ 


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Assume that $(\star)$ holds. The target is smooth, so in particular $\pi\_L(x)$ is a complete intersection. This implies that then (by the condition $(\star$)), $\overline{\pi\_L^{-1}(\pi\_L(x))}$ is also a complete intersection (at least near $x$), but if it is smooth, then it follows that then $X$ has to be smooth near $x$. (This is an iteration of the idea, that if a Cartier divisor is smooth, then so is the ambient space and essentially the same as what Charles suggested.)


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Here is an example that without assuming $(\star)$ the statement is not true.
Let $X\subseteq \mathbb P^3\_{u:v:w:t}$ defined by $uv=w^2$ and let $x=[0:0:0:1]$. Let $L=\{[1:0:0:0]\}$ and $M=(t=0)\subset \mathbb P^3$. Now $(\pi\_L)|\_X$ is dominant onto $M$, so $\overline{\pi\_L(X)}=M\simeq \mathbb P^2\_{v:w:t}$, in particular smooth. Furthermore, $\pi\_L(x)=[0:0:1]$ and hence $\overline{\pi\_L^{-1}(\pi\_L(x))}=\{[\alpha:0:0:\beta]|[\alpha,\beta]\in \mathbb P^1\}\simeq \mathbb P^1$, also smooth. However, $X$ is obviously not smooth at $x$.


*Remark*: One can also see where the above argument breaks down for this example. Since $(\star)$ is not assumed, it does not follows that $\overline{\pi\_L^{-1}(\pi\_L(x))}$ is a complete intersection (and it isn't). 


