Trying to do an exercise from Analysis I, Tao.

edit: The paragraph introducting/defining inverse:

If $f$ is bijective, then for every $y \in Y$, there is exactly one $x$ such that $f(x)=y$ (there is at least one because of surjectivity, and at most one because of injectivity). This value of $x$ is denoted $f^{-1}(y)$; thus $f^{-1}$ is a function from $Y$ to $X$. We call $f^{-1}$ the inverse of $f$.

(The paragraph stands alone, but I assume $f:X \rightarrow Y$)

Exercise 3.3.6.

Let $f:X\rightarrow Y$ be a bijective function, and let $f^{-1}:Y \rightarrow X$ be its inverse. Verify the cancellation laws $f^{-1}(f(x))=x$ for all $x \in X$ and $f(f^{-1}(y))=y$ for all $y \in Y$. Conclude that $f^{-1}$ is also invertible, and has $f$ as its inverse (thus $(f^{-1})^{-1}=f$).

My attempt at a proof:

(1) Let $y \in Y$. Since $f$ is bijective, we have exactly one $x \in X$, denoted $f^{-1}(y)$, such that $y=f(x)=f(f^{-1}(y))$.

(2) Now let $x \in X$. Then $f(x)=y$ for exactly one $y \in Y$. Furthermore, since $f$ is bijective we have exactly one $x_0 \in X$, denoted $f^{-1}(y)$, such that $y=f(x_0)$. Thus $x_0=x$, and so $x=f^{-1}(y)=f^{-1}(f(x))$. Observe that we have shown that for all $x \in X$, there is exactly one $ y \in Y$, namely $y=f(x)$, such that $f^{-1}(y)=x$, thus $f^{-1}$ is invertible and has $f$ as its inverse.


I'm not sure if the proof is correct. I believe that (1) proved the cancellation law $y=f(f^{-1}(y))$.

I believe that (2) proved the cancellation law $x=f^{-1}(f(x))$ and also that $f^{-1}$ is onto and one to one, thus invertible.

I'm not sure if the last claim (that $f$ is the inverse of $f^{-1}$) was properly justified.

  • 1
    $\begingroup$ I must say that actually the exercise makes no sense to me. This because in my optics - if $f^{-1}$ is defined to be the inverse of $f$ then by definition we have $f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$. How is "inverse function" defined in your material? $\endgroup$ – Vera Jul 28 '18 at 14:44
  • $\begingroup$ @Vera I added the paragraph where function inverse was discussed. $\endgroup$ – Ken Tjhia Jul 28 '18 at 14:53

The definition of $f^{-1}$ in the first paragraph is $f(f^{-1}(y)) = y$.

If we let $y = f(x)$ we get $f(f^{-1}(f(x))) = f(x)$. Since $f$ is injective we have $f^{-1}(f(x)) = x$.

To show that $f^{-1}$ is bijective, note that if $f^{-1}(a)= f^{-1}(b)$, then applying $f$ to both sides yields $a=b$, hence $f^{-1}$ is injective. Furthermore, $f^{-1}(f(x)) = x$, hence $f^{-1}$ is surjective.

If $g$ is the inverse of $f^{-1}$, then, as in the first paragraph of the question, we have $f^{-1}(g(y)) = y$, and applying $f$ to both sides yields $g(y) = f(y)$ and so $(f^{-1})^{-1} = f$.


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