A topology is coherent with the subspaces if and only if it is the final topology on the inclusion maps Suppose $X$ is a topological space and $\{X_\alpha\}$ is a family of subspaces whose union is $X$. Show that the topology of $X$ is coherent with the subspaces $\{X_\alpha\}$ if and only if it is the finest topology on $X$ for which all of the inclusion maps $i_\alpha: X_\alpha \hookrightarrow X$ are continuous.
The only if part is easy. However, I am struggling with the if direction. How can I show that if there exists a subset $U$ of $X$ for which $U \cap X_\alpha$ is open in $X_\alpha$, but $U$ is not open in $\mathscr{T}$, the topology on $X$, then it is not the final topology for $i_\alpha$? 
 A: Let $S = \{0,1\}$ be the Sierpisnki space (I will note $1$ for the 'open point'). Now, consider the indicator function of $U$ viewed as a map $\chi_U : X \to S$. Each restriction to $X_\alpha$ is continuous, as
$$
(\chi_U\iota_\alpha)^{-1}(1) = i_\alpha^{-1}(U) = U \cap X_\alpha
$$ 
is open in $X_\alpha$ by hypothesis. However, the set $U = \chi_U^{-1}(1)$ is not open and so $\chi_U$ is not continuous. This contradicts that the family of inclusions is final.
Another approach: define $\tau'$ as the topology generated by the subasis $\tau \cup \{U\}$ with $\tau$ the topology on $X$. This is a strictly finer topology as it contains $U$, and each inclusion $i_\alpha$ is continuous: we only need to check that preimages of subbasic open sets are open, and since $i_\alpha : X_\alpha \to (X,\tau)$ is continuous it suffices to observe that by our assumption
$$
i_\alpha^{-1}(U) = U \cap X_\alpha
$$
is open. Again, this is a contradiction.
A: Suppose $\mathscr T$ is the finest topology for which the inclusion maps $i_{\alpha}$ are continuous. Let $U\subseteq X$ such that $U \cap X_{\alpha} \in \mathscr T$ for all
$\alpha.$ If $U$ is not open in $\mathscr T$ then by defining a new topology $\mathscr T\subseteq \mathscr T_1$ by adding $U$ to $\mathscr T$, we obtain a finer topology such that every $i_{\alpha}$ is continuous. This is a contradiction. 
