Asumme that $\Pi : \Bbb P^n \times \Bbb P^m \rightarrow \Bbb P^m $is a closed map. $X\subset \Bbb P^n$ And $Y\subset P^m$ $f: X\rightarrow Y$ be a morphism of projective varieties.
the diagonal $\Delta (Y) =\{(y,y)\in Y \times Y\}$ is closed in $\Bbb P^m \times \Bbb P^m$
I think that since the diagonal is in algebraic subset, it is closed.
But I am not sure this answer. Please explain why the diagonal is closed. Thank you.
First of all, neither $\Pi$, $X$ or $f$ seem to be relevant to the question whether or not $\Delta(Y)$ is closed.
You have embedded $\mathbb P^m\times\mathbb P^m$ into $\mathbb P^{M}$ with $M=(m+1)^2-1$ via the Segre embedding, presumably. This yields a closed embedding of $Y\times Y$ into $\mathbb P^M$. Under this embedding, a point $([x_0:\ldots:x_m],[x_0:\ldots:x_m])\in\Delta(Y)$ corresponds to the point $$[x_ix_j] = \left[ \begin{matrix} x_0x_0 & \cdots & x_0x_m \\ \vdots & \ddots & \vdots \\ x_mx_0 & \cdots & x_mx_m \end{matrix}\right]$$ I claim that $[z_{ij}] \in \mathbb P^N$ is an element of the diagonal $\Delta(Y)$ if and only if $[z_{ij}]\in Y\times Y$ and $z_{ij}-z_{ji}=0$ for all $i,j$. Since $z_{ij}=z_{ji}$ is a polynomial condition, this would prove that $\Delta(Y)$ is closed.
We have already seen that the "only if" part is satisfied.
For the converse, assume that $[z_{ij}]\in Y\times Y$ satisfies $z_{ij}=z_{ji}$. We can write $[z_{ij}]=[x_iy_j]$ with $[x_0:\ldots:x_n],[y_0:\ldots:y_n]\in Y$, because $[z_{ij}]$ is in the image of the Segre embedding. Without loss of generaliy, we assume $y_0=1$. The condition $z_{ij}=z_{ji}$ means $x_iy_j=x_jy_i$. For $i=0$, this means $x_0y_j=x_j$ for all $j$. Consequently, $x_0$ can not vanish, otherwise we'd have $x_j=0$ for all $j$. Scaling to $x_0=1$ yields $y_j=x_j$. Thus, $[z_{ij}]$ corresponds to a point in the diagonal.