Is the restriction of an inverse function invertible?

Question

Let $f:A\rightarrow B$, where $A$ and $B$ are open sets of $\mathbb{R^n}$ for some n, be invertible. Let $C$ and $D$ be open subsets of $A$ and $B$ respectively. Is $f:C \rightarrow D$ invertible?

Answer 1

One can discuss this question more generally, without restriction to $\mathbb{R}^n$ and open sets:

"Let $$f:A\rightarrow B\tag{1.1}$$ be an invertible function and $C\subset A$ and $D\subset B$. Is the function $$f:C\rightarrow D\tag{1.2}$$ invertible, too?"

I think this question is not well posed. If $f$ is a fuction from $A$ to $B$ it cannot be a function from $C$ to $D$ at the same time, at least from my definiton of a function. So I would wirte the question this way:

Let $$f:A\rightarrow B\tag{2.1}$$ be an invertible function and $C\subset A$ and $D\subset B$. Is the function $g$ invertible, too, where $$g:C\rightarrow D\tag{2.2}$$ and $$g(x)=f(x), \forall x \in C\tag{2.3}$$

This still does not satisfy me, because without further restriction, we cannot guarantee that $(2.2)$ and $(2.3)$ are meaningful definitions, if we do not require that $$f(x) \in D, \forall x \in C$$

So finally I would pose the following question

"Let $$f:A\rightarrow B\tag{3.1}$$ be an invertible function and $C\subset A$ and $D\subset f(A)$. Is the function $g$ invertible, too, where $$g:C\rightarrow D\tag{3.2}$$ where $$g(x)=f(x), \forall x \in C\tag{3.3}$$

Now the $g$ is a correctly defined function.

A function $h:X\rightarrow Y$ is invertible if and only if

• $h$ is injective: $$\forall x_1 \in X\;\forall x_2 \in X\!: h(x_1)=h(x_2) \implies x_1=x_2$$
• $h$ is surjective: $$\forall y \in Y \;\exists x \in X\!: f(x)=y$$

From $$g(x_1)=g(x_2)$$ follows $$f(x_1)=f(x_2)$$ and if $f$ is injective then $$x_1=x_2$$ So if $f$ is injective then $g$ is injective.

If $y \in D$ then there is an $x\in A$ such that $f(x)=y$. But we do not know if $x \in C$. We can find simple examples, where this does not hold, e.g. $$f:x\mapsto x$$ where $A=B=\{1,2,3\}$ and $C=\{1\}, D=\{1,2\}$.Here $g(x)=2$ does not have a solution. But if $D=f(C)$ then $g$ is surjective, too. The requirement that $A,B,C,D$ are open subsets of $\mathbb{R}^n$ and does not change anything.

So finally we have:

"Let $$f:A\rightarrow B$$ be a function and $C\subset A$ and $f(C)\subset D$, and $g$ the function $$g:C\rightarrow D$$ where $$g(x)=f(x), \forall x \in C$$

if $f$ is injective, then $g$ is injective. $g$ is surjective if and only if $D=f(C)$. $g$ is invertible (bijective) if and only if $f(C)=D.$

If $g$ is the restriction of $f$ to $C$, then this is often written as $f|_C$, then this is usually defined as

$$f|_C:C\rightarrow B$$ $$f|_C(x)=f(x),\forall x \in C$$

If $f$ is injective then $f|_C$ is injective. If $f$ is surjective then $f|_C$ must not be surjective.

Answer 2

If you don't put any requirements on what the restrictions $C$ and $D$ are, then the answer is not satisfying. To see why, suppose $f : \mathbf{R} \rightarrow \mathbf{R}$ is the identity map. Let $C = (0,1)$ and $D = (1, \infty)$ be open subsets of the domain and range, respectively. Then the identity map can't even be defined.

If you do not allow for such trivial counterexamples, then the answer is yes. Just use the original inverse for $f$ and restrict it to the new domain. Note that since you merely asked for an inverse mapping, the topology doesn't really play a role, so you don't need the open criterion in your question.