Let $j:X\rightarrow Y$ be a morphism of schemes, i'm trying to prove that $j$ is a monomorphism if and if only the diagonal embedding $\Delta_{X/Y}: X\rightarrow X\times_YX$ is an isomorphism, here is what i've tried so far for the if only part
If $j$ is a monomorphism, it follows that $pr_1=pr_2=p$ and $p \circ \Delta_{X/Y}=id$, i'm clueless of what to do next
Observe that $pr_1=pr_2$ if $\Delta_{X/Y}$ is an isomorphism.
Consider a commutative diagram $\require{AMScd}$ \begin{CD} W @>f>> X\\ @V{g}VV @V{j}VV\\ X @>{j}>>Y \end{CD} We have to show $f=g$
By universal property of fibered product $\exists ! $ morphism $\theta: W\rightarrow X\times_YX$ such that $f=pr_1\circ \theta$ and $g=pr_2\circ \theta$. But $pr_1=pr_2=p$. So $f=g=p\circ \theta $. Thus $j$ is a monomorphism.
Assume $j$ is a monomorphism, then you have $pr_1=pr_2$. So you have two morphisms $\Delta :X \rightarrow X\times_Y X$ and $p:X\times_Y X\rightarrow X$. We also know $p\circ\Delta=Id$
Consider the commutative diagram $\require{AMScd}$ \begin{CD} X\times_Y X @>p>> X\\ @V{p}VV @V{j}VV\\ X @>{j}>>Y \end{CD} Look at $X\times _Y X\xrightarrow{\Delta\circ p} X\times_Y X$ Observe that $pr_1\circ \Delta \circ p=p=pr_2\circ \Delta \circ p$. But $\exists !$ $\theta : X\times_Y X\rightarrow X\times_Y X$ such that $pr_1\circ \theta= pr_2\circ \theta $
This forces $\Delta \circ p=Id $