Vtyjkyuk

Consider $Z : = cup_i in I Z_i$ a disjoint union of topological spaces and $f : Z to A$ a surjective function. Put the quotient topology on A and then prove that a subset $U subseteq A$ is open in $A$ if and only if for all $i in I, f^-1 (U) cap Z_i$ is open in $Z$. Prove that a subset $V subseteq A$ is closed in $A$ if and only if for all $i in I, f^-1 (V) cap Z_i$ is closed in $Z$.

We define the quotient topology on A by $Im_A : = f^-1(U) in Im $ where $Im$ is the disjoint union topology generated from $Z$.

Here is my attempt for the first part.

Assume $U subset A$ is open in $A$. Then $U in Im_A$ which implies that $f^-1(U) in Im $ by definition of $Im_A$. Then we fix $i$. So $$f^-1(U)cap Z_i = V subseteq f^-1(U) subseteq Im_i subseteq Im $$

I am struggling with the reverse direction and proving for closed sets.

$O$ is open in $A$ iff $f^-1[O]$ is open in $Z$ (this is the definition of the quotient topology on $A$)

A set $O$ is open in $Z$ iff for all $iin I$, $O cap Z_i$ is open in $Z_i$. (definition of the disjoint sum topology)

Combining: $O$ is open in $A$ iff for all $i$, $f^-1[O] cap Z_i$ is open in $Z_i$. And note that $f^-1[O]cap Z_i = (f|_Z_i)^-1[O]$ so that we get that all $f$ restricted to $Z_i$ are also quotient maps (where we don't ask quotient maps to be surjective necessarily).

Both the characterisations of the quotient and the sum topology hold with closed sets instead of open sets as well; the proof of this is completely standard involving some simple set theory with taking complements; try it.