# Given the projection, can we determine the object?

Background:

I was recently working on some exercises in geometric probability, in which you visualize the probability space of some event (and also the sample space) as a region in space, and you solve the problem by computing the areas/volumes of the probability space and sample space. In this problem: Variant on classic geometric probability: 3 people meeting during the day, we have three people who each visit some place at random times during the day, and each stay there for an hour, and we would like to find out the probability that at least two meet. In solving it, I needed to figure out the probability that all three people meet. I did this by noting that the probability space of this event forms a region in 3 dimensions, and if projected onto the $xy,yz,xz$ planes gives a region representing the probability space that two given people meet. I solved the two dimensional problem first, then knowing the shapes of the projections, I guessed the shape of the 3D region. I suspect that this problem can be solved by a similar method although we have to work in 4 dimensions now:Probability that two people see eachother TWICE during the day.

So this brings up the question: suppose we are given some closed surface $S$ in 3 dimensional space. Let the projection of $S$ onto the $xy,yz,xz$ planes be $P_1$, $P_2$, $P_3$, respectively. Given $P_1,P_2,P_3$, is $S$ uniquely determined? What about for higher dimmensions?

The answer to the problems seems to be yes in 3 dimensions by the following intuitive argument: Drag $P_1$ through space such that for each point in $P_1$ changes the $z$ coordinate but $x$ and $y$ coordinates remain the same, and do similarly for $P_2$ and $P_3$. You get tree infinite solids $Q_1$, $Q_2$, $Q_3$. Suppose the intersection of all three solid (if it exists) is some region $R$. Assuming the boundary $R$ forms a non-intersecting, closed surface, then the boundary is exactly the surface $S$. The problem with this argument is its not rigorous, and I think I'm making an assumption which is false somewhere.