Tao Analysis I - 3.4.2

Exercise 3.4.2

Let$ f:X \to Y$ be a function from one set X to another set $Y$, let $S$ be a subset of $X$, and let $U$ be a subset of $Y$.

(i) What, in general, can one say about $f^{-1}( f (S))$ and $S$?

(ii) What about $f ( f^{-1}(U))$ and $U$?

(iii) What about $f^{-1}( f ( f^{-1}(U)))$ and $f^{−1}(U)$?

Let's draw a picture to aid our thinking.

The first thing to note is that $f$ hasn't been defined as injective, so we can't assume it is invertible in the sense that it has an inverse function. We can still discuss an inverse image.

The second thing to say is that $S$ and $U$ are not defined as being related, $U$ is not the image of $S$ under $f$.

(i) What, in general, can one say about $f^{-1}( f (S))$ and $S$?

Since $f$ is defined as a function from $X$ to $Y$, and since $S \subseteq X$, then $f(S)$ is defined, and is a subset of $Y$. 

Now, by Definition 3.4.5, $f^{-1}(f(S))$ is the inverse image of $f(S)$ where 

$$f^{-1}(f(S)) = \{x \in X: f(x) \in f(S)\}$$

If $x \in S$, then $f(x) \in f(S)$. This means such $x$ belong in the set $\{x \in X: f(x) \in f(S)\}$, which is $f^{-1}(f(S))$.

So we can say $S \subseteq f^{-1}( f (S))$. $\square$

Note, we can't say $f^{-1}( f (S)) \subseteq S$, and so we can't say the two sets are equal.

(ii) What about $f ( f^{-1}(U))$ and $U$?

We have to be careful here. Every $x \in X$ has an $f(x) \in Y$. However, not every $y \in Y$ has an $x$ such that $f(x) \in Y$. This is an asymmetry in the definitions of domain and codomain.

The most we can say about $U$ is that some $x \in X$, and possibly no $x \in X$, are such that $f(x) \in U$. We are not given any facts that tell us every $y \in Y$ has an antecedent in $X$.

Now, Definition 3.4.5 tell us that

$$f^{-1}(U) = \{x \in X: f(x) \in U\}$$

In words, this inverse image is a set of $x$ such that $f(x) \in U$. So if we apply $f$ to this set we have

$$f(f^{-1}(U) \subseteq U$$

We can say $f ( f^{-1}(U)) \subseteq U$, where that subset could be the empty set. $\square$

We can't say the sets are equal because there may be only some, or no, $y \in U$ such that $f(x) = y$.

(iii) What about $f^{-1}( f ( f^{-1}(U)))$ and $f ^{−1}(U)$?

We can use the result from (i), $S \subseteq f^{-1}( f (S))$ and set $S=f ^{−1}(U)$ to give,

$$f ^{−1}(U) \subseteq f^{-1}( f (f ^{−1}(U)))$$

Now let's use the result from (ii), $f(f^{-1}(U) \subseteq U$ and take the inverse image of both sides to give,

$$f^{-1}(f(f^{-1}(U)) \subseteq f^{-1}(U)$$

This uses the fact that $A \subseteq B \implies f^{-1}(A) \subseteq f^{-1}(B)$, see below for proof.

By showing inclusion in both directions, we have shown the two sets are equal, $f^{-1}( f ( f^{-1}(U))) = f ^{−1}(U)$. $\square$

Show $A \subseteq B \implies f^{-1}(A) \subseteq f^{-1}(B)$

The definition of inverse image $𝑓^{−1}$ applied to $A$ is

$$f^{-1}(A) = \{x \in X: f(x) \in A\}$$

Similarly for $B$

$$f^{-1}(B) = \{x \in X: f(x) \in B\}$$

Since $A \subseteq B$, then $f(x) \in A \implies f(x) \in B$.

Thus,

$$f^{-1}(A) \subseteq f^{-1}(B) \; \square$$