Answer
\(f\) is **one‐to‐one** and **onto**.
\(g\) is also a bijection.
\(g = f^{-1}\).
No counterexample exists because the hypotheses \(g\circ f = I_X\) and \(f\circ g = I_Y\) precisely characterize \(f\) as a bijection with inverse \(g\).
Work Step by Step
**Short Answer:** The statement is **true**. There is no counterexample. If \(f : X \to Y\) and \(g : Y \to X\) satisfy
\[
g \circ f \;=\; I_X
\quad\text{and}\quad
f \circ g \;=\; I_Y,
\]
then \(f\) and \(g\) are indeed **bijections**, and \(g\) is precisely the inverse function \(f^{-1}\).
Below is a concise proof.
---
## 1. Show \(f\) is One-to-One (Injective)
Suppose \(f(x_1) = f(x_2)\) for some \(x_1, x_2 \in X\). Then
\[
x_1
\;=\;
I_X(x_1)
\;=\;
\bigl(g \circ f\bigr)(x_1)
\;=\;
g\bigl(f(x_1)\bigr)
\;=\;
g\bigl(f(x_2)\bigr)
\;=\;
\bigl(g \circ f\bigr)(x_2)
\;=\;
I_X(x_2)
\;=\;
x_2.
\]
Hence \(x_1 = x_2\). This proves \(f\) is injective.
---
## 2. Show \(f\) is Onto (Surjective)
Take any \(y \in Y\). Then
\[
y
\;=\;
I_Y(y)
\;=\;
\bigl(f \circ g\bigr)(y)
\;=\;
f\bigl(g(y)\bigr).
\]
So \(y\) is in the image of \(f\). Because \(y\in Y\) was arbitrary, \(f\) is surjective.
---
## 3. Conclude \(f\) is a Bijection
Combining injectivity and surjectivity shows \(f\) is a **bijection**. Bijections always have an inverse function.
---
## 4. Identify \(g\) as \(f^{-1}\)
Since \(f\) is a bijection, it has a unique two‐sided inverse \(f^{-1} : Y \to X\) satisfying
\[
f^{-1}\circ f = I_X
\quad\text{and}\quad
f \circ f^{-1} = I_Y.
\]
But \(g\) already plays that exact role:
\[
g \circ f = I_X
\quad\text{and}\quad
f \circ g = I_Y.
\]
Hence \(g\) **is** \(f^{-1}\). In particular, \(g\) is also a bijection (since an inverse of a bijection must itself be a bijection).
