# Let A 1 </msub> , A 2 </msub> , P be CPOs and let &#x03C8;<!-- ψ -->

Let ${A}_{1}$, ${A}_{2}$, P be CPOs and let $\psi :{A}_{1}×{A}_{2}\to P$ be a map, then $\psi$ is continuous $\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$ it is so in each variable separately
Proving this $⇒$ direction is quite easy, but I am not sure how to conclude the opposite $⇐$ direction:
If $\bigvee D=\left({d}_{1},{d}_{2}\right)=\bigvee \left({D}_{1}×{D}_{2}\right)$ , we have:
$\psi \left(\bigvee D\right)=\psi \left({d}_{1},{d}_{2}\right)={\psi }^{{d}_{1}}\left({d}_{2}\right)={\psi }_{{d}_{2}}\left({d}_{1}\right)$ , where
${\psi }^{{d}_{1}}\left({d}_{2}\right)={\psi }^{\bigvee {D}_{1}}\left(\bigvee {D}_{2}\right)=\bigvee {\psi }^{\bigvee {D}_{1}}\left({D}_{2}\right)=\bigvee \psi \left(\left\{\bigvee {D}_{1}\right\}×{D}_{2}\right)$
and, similarly,
${\psi }_{{d}_{2}}\left({d}_{1}\right)={\psi }_{\bigvee {D}_{2}}\left(\bigvee {D}_{1}\right)=\bigvee {\psi }_{\bigvee {D}_{2}}\left({D}_{1}\right)=\bigvee \psi \left({D}_{1}×\left\{\bigvee {D}_{2}\right\}\right)$
how to conclude that $\psi \left(\bigvee D\right)=\bigvee \psi \left(D\right)$ ?
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Trey Ross
Step 1
Anyway, I'll start in a point you have already reached.
$\begin{array}{rl}\psi \left(\bigvee D\right)& ={\psi }_{{d}_{2}}\left({d}_{1}\right)={\psi }_{{d}_{2}}\left(\bigvee {D}_{1}\right)\\ & =\bigvee {\psi }_{{d}_{2}}\left({D}_{1}\right)=\bigvee \left\{{\psi }_{{d}_{2}}\left(x\right):x\in {D}_{1}\right\}\\ & =\bigvee \left\{\psi \left(x,{d}_{2}\right):x\in {D}_{1}\right\}\\ & =\bigvee \left\{{\psi }^{x}\left(\bigvee {D}_{2}\right):x\in {D}_{1}\right\}\\ & =\bigvee \left\{\bigvee \left\{{\psi }^{x}\left(y\right):y\in {D}_{2}\right\}:x\in {D}_{1}\right\}\\ & =\bigvee \left\{\psi \left(x,y\right):x\in {D}_{1},y\in {D}_{2}\right\}\\ & =\bigvee \psi \left(D\right).\end{array}$