Prove $f$ is continuous

Question

Let $X$ be a topology space and let $S_1,...,S_n\subseteq X$ close sets in $X$ such that $S_1\cup...\cup S_n=X$

Let $Y$ be a topology space and let $f:X\to Y$ a function such that $f:S_{|S_k}\to Y$ continuous for all $1\leq k \leq n$

Prove: $f$ is continuous

I tried to show $f$ sends a close set to a close set, to do so I tried to look at the pre-image of each $f(S_k)$ and shoe it is closed but how can I do it?

Answer 1

Let $C$ be closed in $Y$. Then $f|_{S_k} ^{-1} (C)=S_k \cap f^{-1} (C) $ is closed for each $k$. From this we see that $ f^{-1}(C)\equiv \cup_{k=1}^{n} (S_k \cap f^{-1}(C))$ is closed. Since inverse image of every clased set is closed it follows that $f$ is continuous.

Answer 2

For $k=1,2,\dots,n$ let $i_k:S_k\to X$ denote the inclusion.

Then $f\circ i_k:S_k\to Y$ is a continuous function for every $k$.

Now let $F\subseteq Y$ be a closed set.

Then $f^{-1}(F)\cap S_k=i_k^{-1}(f^{-1}(F))=(f\circ i_k)^{-1}(F)$ is closed in $S_k$ for every $k$ where $S_k$ is equipped with the subspace topology inherited from $X$.

That means (for every $k$) that $f^{-1}(F)\cap S_k=F_k\cap S_k$ where $F_k$ is a closed subset of $X$.

But just like $F_k$ also $S_k$ is a closed subset of $X$ so we conclude that $f^{-1}(F)\cap S_k=F_k\cap S_k$ is a closed subset of $X$ (as intersection of closed sets).

Then it follows that $f^{-1}(F)=\bigcup_{k=1}^n(f^{-1}(F)\cap S_k)$ is a closed subset of $X$ (as finite union of closed sets).

Proved is not that preimages of closed subsets of $Y$ under $f$ are closed subsets of $X$, or equivalently that $f$ is continuous.

Answer 3

$\textbf{Definition}$. We say that $f:X\to Y$ is continuous in a point $x_0\in X$ if for each neighbourhood $V_{f(x_0)}$ of $f(x_0)$ the inverse image of this neighbourhood, $f^{-1}(V_{f(x_0)})$, is a neighbourhood of $x_0$.

We can generalize the previous definition saying that $f:X\to Y$ is continuous in $A\subset X$ if it's continuous in each point of $A$.

Firstly, we're going to prove that $f$ is continuous iff the inverse image of an open set is open

$\boxed{\Rightarrow}$ Let $U$ be an open set of $Y$, so it's a neighbourhood of each of its points. However, for each $f(x_0)\in U$, as $f$ is continuous, the inverse image of $U$ is a neighbourhood of $x_0$. Furthermore, for each $x_0\in f^{-1}(U)$ it's verified that $f^{-1}(U)$ is a neighbourhood of $x_0$, so it's a neighbourhood of each point, that is to say, it's open.

$\boxed{\Leftarrow}$ Let $V_{f(x_0)}$ be a neighbourhood of $f(x_0)$ in $Y$. By definition, it contains and open set $U$ such that $f(x_0)\in U$. Now, by hypothesis, $f^{-1}(U)$ is open. But we also have that $x_0\in f^{-1}(U)\subset f^{-1}(V_{f(x_0)})$, so $f^{-1}(V_{f(x_0)})$ contains an open set which contains $x_0$. Thus, it's a neighbourhood and $f$ is continuous because we've done this for an arbitrary point.

Now, we're goint to prove that the inverse image of an open set is open iff the inverse image of an closed set is closed. Let $F\subset Y$ be a close set, that is to say, $Y\backslash F$ is open. By hypothesis, $f^{-1}(Y\backslash F)$ is open, and $$ f^{-1}(Y\backslash F)=X\backslash f^{-1}(F)$$ so $f^{-1}(F)$ is close. The other implication is analogue.

Finally, we're going to check that the inverse image of an closed set is closed iff there is a finite cover of $X$ $$X=\bigcup_{i=1}^k F_i$$ that verifies that every restriction $f|_{F_i}$ is continuous.

$\boxed{\Rightarrow}$ We have that $f$ is continuous because the inverse image of an closed set is closed (we've already proved that). Taking as the cover the set $\{X\}$ we're finished.

$\boxed{\Leftarrow}$ Let $F\subset Y$ a closed set. Then, we can write $$f^{-1}(F)=f^{-1}(F) \cap X= f^{-1}(F) \cap \bigcup_{i=1}^k F_i=\bigcup_{i=1}^k f^{-1}(F)\cap F_i = \bigcup_{i=1}^k (f|_{F_i})^{-1}(F).$$ However, $f|_{F_i}$ is continuous, so $(f|_{F_i})^{-1}(F)$ are closed in $F_i$. By relative topology, $(f|_{F_i})^{-1}(F)$ is closed in $X$ too, ando the inverse image of $F$ is closed because it's a finite union of closed sets.