Suppose $(X_i)_i$ is an indexed family of non-empty topological spaces. Recall: $\coprod_{i\in I}X_i = $ $\{$ $(x,i)$ $:$ $x\in X_i$ and $i\in I$ $\}$ . There is a canonical injection $\sigma_i: X_i \rightarrow \coprod_{i\in I}X_i$ , given by $\sigma_i(x)=(x,i)$. We usually identify each set $X_i$ with its image, $X_i^*=\sigma_i(X_i)$.
Show that:
If $U$ is open in $X_{i_0}^*$ then $U$ is open in $X_{i_0}$.
My attempt:
Observe that if $U$ is open in $X_{i_0}^*$ then $U$ is open in the disjoint union, and so by definition of the topology on the disjoint union, $\sigma_{i_0}^{-1}(U)$ is open in $X_{i_0}$.
Is that all that is required?
This is trivially true "by your identification". Note that if $U$ is a subset of $X^\ast_{i_0}$ it is by definition not a subset of $X_{i_0}$. So in that sense the statement has no meaning. But we identify $X^\ast_{i_0}$ via the embedding $\sigma_{i_0}$ so for every fixed $i \in I$ and every $U \subseteq X_i$ we have $$U \text{open in } X_i \iff \sigma_i[U] \text{ open in } X^\ast_i$$
and likewise $$U \subseteq X_i^\ast \text{ open } \iff \sigma_i^{-1}[U] \text{ open in } X_i$$
Both statements are almost trivial consequences of the fact that $\coprod_{i \in I} X_i$ has the final topology with respect to the $\sigma_i, i \in I$, and the definition of the subspace topology (of the $X_i^\ast$ relative to $\coprod_{i \in I} X_i$).