i}x_i$ is of the form $u = \coprod_{j \in i}u_j^*$ where $u_j^*$ is open in $x_j^*$ for all $j \in i$. but now $$\sigma^{ }_i(u) = \sigma_i^{ }(\coprod_{j \in i}u_j^*) = \sigma^{ }_i(u_i^*) = u_i$$ share | cite | improve this answer answered mins ago olivier roche olivier roche silver badges bronze...
share | improve this answer answered hours ago rsf rsf gold badge silver badges bronze badges my experience matches shivam and over the past years, jobs and interviews it is exactly what i've experienced. i just left a place that hired sdets as part of their 'agile trnaformation'. most of them ended...