Tao Analysis I - 3.4.1

Exercise 3.4.1

Let $f : X \to Y$ be a bijective function, and let $f^{−1} : Y \to X$ be its inverse. Let $V$ be any subset of $Y$. Prove that the forward image of $V$ under $f^{−1}$ is the same set as the inverse image of $V$ under $f$; thus the fact that both sets are denoted by $f^{−1}(V)$ will not lead to any inconsistency.

There are two, sufficiently different, definitions of inverse; that of an inverse function, and the inverse image. Let's remind ourselves of the difference.

Inverse function. If $f:X \to Y$ is bijective, then for every $y \in Y$, there is exactly one $x \in X$ such that $f(x)=y$. This value is denoted $f^{-1}(y)$, where $f^{-1} : Y \to X$ is the inverse of $f$.

Definition 3.4.5 (Inverse images). If $U$ is a subset of $Y$, we define the set $f^{-1}(U)$ to be the set

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

In other words, $f^{-1}(U)$ consists of all the elements of $X$ which map into $U$:

$$f(x) \in U \iff x \in f^{-1}(U)$$

We call $f^{-1}(U)$ the inverse image of $U$.

The key difference is that inverse images don't require a function to be bijective. For example, $f:\mathbb{Z} \to \mathbb{N}$ where $f(x)=x^2$ is not bijective because both $f(2)=4$, and $f(-2)=4$. This function has no inverse, but we can define an inverse image. Consider $U=\{0, 1, 4\}$, then the inverse image $f^{-1}(U)=\{-2, -1, 0, 1, 2\}$.

Let's proceed with the exercise by starting with a picture to help us think.

We need to show that the following two sets are the same:

• the image of $V$ under the function $f^{-1}$
• the inverse image of $V$ under $f$

Let's start with the image $f^{-1}(V)$. This is the set $\{f^{-1}(v): v \in V\}$. Now, because $f$ is bijective, it is invertible. By definition of inverse functions, for every $v \in V$, there is a single $x = f^{-1}(v)$ such that $f(x)=v$. We can therefore substitute $f^{-1}(v)$ with $x$, and $v$ with $f(x)$ in the set definition to give $\{x: f(x) \in V\}$.

By Definition 3.4.5, the inverse image of $V$ under $f$ is that set of $x$ such that $f(x) \in V$. This set is defined as $\{x: f(x) \in V\}$.

The two sets have the same membership condition, and so are the same. $\square$

Note that this equivalence between the two definitions hinges on $f$ being bijective. If $f$ were not bijective, then the two sets are not necessarily the same.

Alternative Solution

ProFatXuan's solution takes the standard approach of proving two sets, $A$ and $B$, are the same by showing that an element of one is an element of the other, $A \subseteq B$, and $B \subseteq A$. Let's develop a proof using that approach.

Let's define the two sets, $A$ and $B$:

• $A$ is the forward image of $V$ under $f^{-1}$. So $A=\{f^{-1}(v): v\in V\}$
• $B$ is the inverse image of $V$ under $f$. So $B=\{x \in X: f(x) \in V\}$

Now, if $x \in A$ then $x=f^{-1}(v)$ for $v \in V$. By definition of an inverse function, applicable since $f$ is bijective, every $v \in V$ has a single $x=f^{-1}(v)$ such that $f(x)=v$. This is the definition of $B$ so $A \subseteq B$.

If $x \in B$, then it is because those $x$ are such that $f(x)=v$ for $v \in V$. Again, by definition of an inverse function, we have $x=f^{-1}(v)$ for $v \in V$. This is the definition of $A$, so $B \subseteq A$.

By showing elements of one set are members of the other, we have proved the sets are the same. $\square$