Answer
See explanation
Work Step by Step
**Statement to Prove:**
If \(f : X \to Y\) is a one‐to‐one and onto function (i.e., a bijection) with inverse \(f^{-1} : Y \to X\), then
\[
f \circ f^{-1} \;=\; I_Y,
\]
where \(I_Y\) is the identity function on \(Y\). In other words, for every \(y \in Y\), \((f \circ f^{-1})(y) = y\).
---
## 1. Setup and Definitions
1. **Bijection**: Since \(f\) is one‐to‐one and onto:
- Onto means: For each \(y \in Y\), there exists an \(x \in X\) such that \(f(x) = y\).
- One‐to‐one means: If \(f(x_1) = f(x_2)\), then \(x_1 = x_2\).
2. **Inverse** \(f^{-1} : Y \to X\) satisfies:
- For every \(x \in X\), \(f^{-1}\bigl(f(x)\bigr) = x\).
- For every \(y \in Y\), \(f\bigl(f^{-1}(y)\bigr) = y\).
3. **Identity Function** \(I_Y : Y \to Y\) is defined by \(I_Y(y) = y\) for all \(y \in Y\).
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## 2. Proof
Let \(y \in Y\) be arbitrary. We want to show that \((f \circ f^{-1})(y) = y\).
1. **Apply** \(f^{-1}\) to \(y\): let \(x = f^{-1}(y)\). Then \(x \in X\).
2. **Apply** \(f\) to \(x\): by the definition of inverse, \(f(x) = y\). More explicitly, since \(x = f^{-1}(y)\), we have \(f\bigl(f^{-1}(y)\bigr) = y\).
3. **Hence**
\[
(f \circ f^{-1})(y)
\;=\; f\bigl(f^{-1}(y)\bigr)
\;=\; y.
\]
Because \(y\) was an arbitrary element of \(Y\), we conclude that for **all** \(y \in Y\), \((f \circ f^{-1})(y) = y\). Therefore,
\[
f \circ f^{-1} = I_Y.
\]
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### Conclusion
This shows precisely that composing \(f^{-1}\) on the left by \(f\) yields the identity on \(Y\). Hence, \(f \circ f^{-1} = I_Y\), completing the proof of Theorem 7.3.2(b).
