Is the categorical product for projective spaces essentially the tensor product? I wonder whether the categorical product of two projective spaces is essentially given by the tensor product of the underlying vector spaces. Is this at least true for projective Hilbert spaces?
One problem I have with verifying this gut feeling is that I don't even know which morphisms are allowed between two projective spaces. Every non-zero linear map between the underlying vector spaces gives rise to a projective morphism between the projective spaces. But is every projective morphism of this form, or are there other possibilities? What about inversions, for example? (For the Hilbert space case, are the projective morphisms induced by the continuous linear maps?)
 A: The question only makes sense when you specify the category in which you are working.
For an algebraic geometer the most natural choice would be the category of varieties (or even better, schemes) over some field $k$. Then the product $\mathbb{P}(V) \times \mathbb{P}(W)$ embeds into $\mathbb{P}(V \otimes W)$, via the Segre embedding mapping a pair of lines $L \subseteq V$, $L' \subseteq W$ to the line $L \otimes L' \subseteq V \otimes W$. You can explicitly write down the equations which cut out the image, which is known as the Segre variety.
But that $\mathbb{P}(V) \times \mathbb{P}(W)$ differs from $\mathbb{P}(V \otimes W)$ can already be seen by comparing the dimensions. Since $\dim(\mathbb{P}(V))=\dim(V)-1$, we have $\dim(\mathbb{P}(V) \times \mathbb{P}(W)) = \dim(V)+\dim(V)-2$, whereas $\mathbb{P}(V \otimes W) = \dim(V) \dim(W)-1$.
A: A categorical product $X\times Y$ must satisfy the universal property that for each object $Z$ and morphisms $f$ and $g$, there exists a unique morphism $h$ such that the following diagram commutes
$$\begin{matrix}
& &Z\\
&\swarrow^f&\downarrow^h&\searrow^g\\
X&\leftarrow^p&X\times Y&\rightarrow^q&Y
\end{matrix}$$
If we want to show that $X\otimes Y$ is isomorphic to $X\times Y$, the first thing is to define the projections $p$ and $q$. Let $k=(k_0,k_x,k_y,k_{xy})$ be a representative in $X\otimes Y$. We define $p$ and $q$ by the representatives $p(k)=(k_0,k_x)$ in $X$ and $q(k)=(k_0,k_y)$ in $Y$. The problem is that $p$ is not defined for the points $k$ with $k_0=0$ and $k_x=0$, and similar for $q$. But our category might contain a suitable subobject of $X\otimes Y$ excluding these problematic points, so let's postpone this problem.
The other thing we need to do is to define the morphism $h$ for given morphisms $f$ and $g$, and show that it is unique. Let $f(z)=(f_0,f_x)$ and $g(z)=(g_0,g_y)$ be representatives in $X$ and $Y$. We define $h(z)$ to be represented by $(f_0g_0,f_xg_0,f_0g_y,f_x \times g_y)$. The uniqueness of the first three components is easy to see. In order for the fourth component to also be unique, the allowed morphisms in our category have to be sufficiently limited. (For example if $h$ is allowed to let the images of $z$ with $f_0(z)=g_0(z)=0$ undefined, we could use $0$ for the last component. But then the last component would be superfluous, hence would have to be absent in the categorical product.) So let's discuss the issue with the allowed morphisms.

For the answer by Martin Brandenburg, the categorical product is given by a Segre variety, which is a subobject of $X \otimes Y$ by the Segre embedding of dimension $\dim X+\dim Y$. The dimension of $X \otimes Y$ is $(\dim X+1)(\dim Y+1)-1$, so the only times when it is isomorphic to $X \times Y$ is for $\dim X=0$ or $\dim Y=0$.
The question suggested that the non-zero linear maps would be nice as allowed morphisms. However, the composition of two non-zero linear maps doesn't need to be non-zero, so this notion of morphism is problematic. In addition, the map $h$ from above is not really induced by a non-zero linear map. So the answer to the initial question is that the categorical product of projective spaces is related to their tensor product, but isn't isomorphic to it in any easily definable precise way.
