BJT common-base output resistance derivation
The upper current is indeed \$g_mv_\pi\$ downward. This is the collector current flowing from collector to emitter. But since \$v_\pi=-v\$, it'll become \$g_mv\$ upward. The lower current \$\frac{v}{r_e}\$ or \$g_mv\$ must be up so that it cancels out the \$g_mv\$ portion of the \$r_o\$ current - because the same current coming into the collector, meaning \$i_x\$, must come out at the other side of it.

Also this diagram only shows the collector current at the emitter. It doesn't show the contribution from base current at the emitter. So it might be a little bit confusing at the first glance.