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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)

  • What about $f:[0,1)\rightarrow\mathbb{R}$ and $g:[1,2]\rightarrow \mathbb{R}$ defined by $f(x)=0$ for all $x \in [0,1)$ and $g(x)=x$ for all $x \in [1,2]$? The individual functions are continuous over their domains, but the combined function is discontinuous over the interval $[0,2]$. – Michael Jan 01 '20 at 21:47
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    @Michael The topology on your union is not the disjoint union (or coproduct) topology though. If you did take that topology, the combined function would be continuous. – paul blart math cop Jan 01 '20 at 21:52
  • @Michael that's not a disjoint union/coproduct topology on $[0,2]$. – Henno Brandsma Jan 01 '20 at 22:34
  • @Henno Brandsma In the disjoint union topology, all summands most be open, but $[1, 2]$ is not open in $[0, 2]$, so this is not the topological coproduct (which is what is meant by disjoint union) of $[0, 1)$ and $[1, 2]$. – paul blart math cop Jan 01 '20 at 22:38
  • @Hennno Brandsma No problem – paul blart math cop Jan 01 '20 at 22:43
  • @HennoBrandsma : You are allowed to delete redundant comments. – Michael Jan 02 '20 at 01:31

1 Answers1

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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.

Henno Brandsma
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  • I dont think my last sentence is correct though, because if $f^{-1}(U) \cap X_j$ is open in $X_j$ , its equal to $U’ \cap X_j$ where U is open in the disjoint union that doesnt mean the preimage of U is equal to U’, which is what i was thinking at the time. But your last sentence fixes that, I think. –  Jan 02 '20 at 08:52
  • Hello, may you please elaborate on the converse of the theorem? I would very much like to see how you would approach it, please. –  Jan 05 '20 at 13:37
  • @topologicalmagician the link in the post (long!) has a proof. – Henno Brandsma Jan 05 '20 at 13:38