Suppose $(X_j)_{j\in J}$ is an indexed family of topological space and $Y$ is a topological space.
$f: \coprod _{j\in J}X_j$ $\rightarrow Y$ is continuous $\iff$ The restriction of $f$ to each $X_j$ is continuous.
My Attempt:
Suppose $f$ is continuous. Let j$\in J$ and $U$ be open in $Y$. Then $f|_{X_j}^{-1}(U)=f^{-1}(U)\cap X_j$ is open in $Y$, since $f^{-1}(U)$ is open in the disjoint union space.
For the converse, let $V$ be open in $Y$. Since the restriction of $f$ to each $X_j$ is continuous, it follows that for each $j\in J$, $f|_{X_j}^{-1}(V)=f^{-1}(V)\cap X_j$ is open in $X_j$ which happens if and only $f^{-1}(V)$ is open in the disjoint union.
Is the proof correct? (Please answer this question)
Best Answer
Yes, the proof is correct. It follows directly, as you say, from the characterisation of open sets in $\coprod_{j \in J} X_j$:
$$ O \subseteq \coprod_{j \in J} X_j \text{ open } \iff \forall j \in J: O \cap X_j \text { open in } X_j$$
which is in itself a consequence of the fact that $\coprod_{j \in J} X_j$ has the final topology w.r.t. the set of standard inclusion-embeddings $\{e_j: X_j \to \coprod_{j \in J} X_j \}$. It's in fact an instance of what I call here the universal theorem of continuity of final topologies, which characterises final topologies. Note that $f \circ e_j = f\restriction_{X_j}$, in essence.