# An exercise about the Pasting Lemma.

I have difficulty with the following exercise from Introduction to Topology (by Tej Bahadur Singh)(Exercise 9 on p. 36):

Let $$f: X\to Y$$ be a function between topological spaces, and assume that $$A\cup B= X$$, where $$A-B\subseteq A^\circ$$, and $$B-A\subseteq B^\circ$$. If $$f|_{A}$$ and $$f|_{B}$$ (endowed with the relative topologies) are continuous, show that $$f$$ is continuous.

I tried to use the following facts (from the book mentioned above):

Definition (locally finite). A family $$\{A_i\}$$ of subsets of a space $$X$$ is called locally finite if each point of $$X$$ has a neighborhood $$U$$ such that $$U\cap A_i\neq \varnothing$$ for at most finitely many indices $$i$$.

(1) Let $$\{U_\alpha\}$$ be a family of open subsets of a space $$X$$ with $$X = \bigcup_\alpha U_\alpha$$. Then a function $$f$$ from $$X$$ into a space $$Y$$ is continuous if and only if $$f|_{U_\alpha}$$ is continuous for each index $$\alpha$$. (See Exercise 8 on p. 36)

(2) If a space $$X$$ is the union of a locally finite family $$\{A_i\}$$ of closed sets, then a function $$f$$ from $$X$$ to a space $$Y$$ is continuous if and only if the restriction of $$f$$ to each $$A_i$$ is continuous. (See Corollary 2.1.10 on p. 33)

Using the fact (1), we obtain that $$f|_{A^\circ\cup B^\circ}$$ is continuous. Clearly, $$f|_{A\cap B}$$ is continuous. Since $$A-B\subseteq A^\circ$$ and $$B-A\subseteq B^\circ$$, $$A^\circ \cup B^\circ \cup (A\cap B)= X$$. But $$A\cap B$$ is not open, so I cannot use the fact (1) again, I have not idea what to do next. Any ideas would be appreciated.

Of course $$f$$ is continuous in all points of $$A^° \cup B^°$$. Since $$A^\circ \cup B^\circ \cup (A\cap B)= X$$, it remains to show that $$f$$ is continuous in all points of $$A \cap B$$.

Let $$x \in A \cap B$$ and $$V$$ be an open neighborhood of $$f(x)$$ in $$Y$$. There exist open neighborhoods $$U_A$$ of $$x$$ in $$A$$ and $$U_B$$ of $$x$$ in $$B$$ such that $$f(U_A) \subset V$$ and $$f(U_B) \subset V$$. Choose open $$W_A, W_B \subset X$$ such that $$W_A \cap A = U_A, W_B \cap B = U_B$$. Define $$W = W_A \cap W_B$$. Then $$W$$ is an open neighborhood of $$x$$ in $$X$$. We have $$f(W) \subset V$$: Let $$y \in W$$. But $$y \in A$$ or $$y \in B$$, w.l.o.g. $$y \in A$$. Thus $$y \in W \cap A \subset W_A \cap A = U_A$$ and therefore $$f(y) \in f(U_A) \subset V$$.

• Thank you very much. Aug 1 at 13:01

Maybe using local continuity arguments will do it: if $$x \in X=A \cup B$$. Let $$V$$ be an open neighbourhood of $$f(x)$$.

If $$x \in A^\circ$$ then we can find an open neighbourhood $$U$$ of $$x$$ such that $$U \subseteq A^\circ$$ and $$f[U] \subseteq V$$. This follows from continuity of $$f\restriction_A$$ and the fact that an $$A$$-open subset of $$A^\circ$$ is open in $$X$$ too.

If $$x \in B^\circ$$ we’re done in the same way, mutatis mutandis.

And if $$x \in A \cap B$$ (but in neither interior), we find an $$A$$-open $$U_1 = U’_1 \cap A$$ (so $$U’_1$$ is open in $$X$$) and a $$B$$-open $$U_2 = U’_2 \cap B$$ (ditto) such that $$f[U_1] \subseteq V$$ and $$f[U_2] \subseteq V$$. Then we could hope that $$f[U_1’ \cap V_1’] \subseteq V$$ as well (though I don’t see yet why this would hold if we don’t somehow restrict the larger open subsets in $$X$$…).

Just a thought that might help. Or maybe use nets: if $$x_i \to x$$ then either the net is eventually in one of the interiors or it’s frequently in $$A \cap B$$ and a subnet in $$A \cap B$$ converges to $$x$$ and the image net to $$f(x)$$, but again we lose control over the other points again…

• Thank you, your ideas are really useful. Using local continuity arguments, I have solved it. Aug 1 at 11:53
• @GeorgeBrown can you write your proof in a separate answer? Aug 1 at 12:23
• @GeorgeBrown indeed Paul showed the missing part for the points in $A\cap B$. I stopped to soon. Aug 1 at 12:40
• @GabrielRomon sorry, I found something wrong with my solution. Aug 1 at 12:45