Munkres Section 18 Theorem 18.2(e)

Question

Munkres is proving the following statement:

"Let $f: \ X \ \rightarrow \ Y$ be continuous. If $f(X) \ \subset \ Z \ \subset \ Y$, we show that the function $g: \ X \ \rightarrow \ Z$ obtained from $f$ is continuous."

He proves it the following way:

"Let $B$ be open in $Z$. Then $B \ = \ U \ \cap \ Z$ for some open set $U$ of $Y$. Because $Z$ contains the entire image set $f(X)$, then $f^{-1}(U) \ = \ g^{-1}(B)$. This then implies that $g$ is continuous.

Now, I understand most of this proof, however, I don't understand the how Munkres arrived at $f^{-1}(U) \ = \ g^{-1}(B)$. We get that $g^{-1}(U \ \cap \ Z) \ = \ g^{-1}(U) \ \cap \ g^{-1}(Z) \ = \ g^{-1}(U) \ \cap \ X \ = \ g^{-1}(U)$, but how can we instantly equate this to $f^{-1}(U)$? If we chose a $U$ that is not in $Z$, how would $f^{-1}(U) \ = \ g^{-1}(U)$?

Answer 1

Note that, if $x\in X$,\begin{align}x\in g^{-1}(B)&\iff g(x)\in B\\&\iff g(x)\in U\cap Z\\&\iff f(x)\in U,\end{align}since $f(x)=g(x)$ and since $Z\supset f(X)$.

Answer 2

First, note that $g^{-1}(U)$ isn’t necessarily defined, since the codomain of $g$ is $Z$, and $U$ is an arbitrary open set of $Y$.

To prove this, we see that we can decompose $U$ into pieces: $U = B \cup (Z\setminus U)$, where the preimage (under $f$) of $Z\setminus U$ must be empty. Now with that in hand, we have:

$$f^{-1}(U) = f^{-1}(B\cup (Z\setminus U) = f^{-1}(B) \cup f^{-1}(Z\setminus U) = g^{-1}(B) \cup\emptyset = g^{-1}(B).$$

Answer 3

For all $x \in X$ we have $g(x)=f(x)$ (because we have only changed the formal codomain of $f$ and renamed it $g$.

$f^{-1}(U) = \{x \in X: f(x) \in U\}$ by definition. If $x \in f^{-1}(U)$ we thus know that $f(x) \in U$ and because $f(x) \in f(X) \subset Z$, we know that $f(x) \in Z$. So $f(x)\in U \cap Z = B$, and as $g(x)=f(x)$ for all $x$, by definition we now also know that $x \in g^{-1}(B)=\{x \in X: g(x) \in B\}$. So $f^{-1}(U) \subset g^{-1}(B)$.

If on the other hand $x \in g^{-1}(B)$, then $g(x) \in B$ so $f(x) (=g(x)) \in U$ and $f(x) \in Z$ (really superfluous). So again by definition $x \in f^{-1}(U)$. And so the other inclusion $g^{-1}(B) \subset f^{-1}(U)$ also holds and we have equality.

If you think about it it's a complete triviality. But it shows that an open subset of $Z$ under $g$ has in inverse image that is open quite explicitly, using continuity of $f$.

Answer 4

Correct me if wrong :

Need to show that $g^{-1} (B)$ is open for $B \subset Z$, $B$ open in $Z.$

$B$ open in $Z$ $\iff$ $U \subset Y$, open in $Y$ such that $B=U \cap Z$.

$g^{-1}(B)= f^{-1}(U\cap Z)=$

$f^{-1}(U) \cap f^{-1}(Z)=$

$f^{-1}(U) \cap f^{-1}(Y)=$

$ f^{-1}(U) \cap X= f^{-1}(U)$.

Since $f^{-1}(U)$ is open , $g^{-1}(B)$ is open,

i.e. $g$ continuous.