Answer
See explanation
Work Step by Step
We want to show that if \(f: X \to Y\) is any function and \(I_Y: Y \to Y\) is the identity function on \(Y\) (so \(I_Y(y) = y\) for every \(y \in Y\)), then
\[
I_Y \circ f \;=\; f.
\]
---
## 1. Definitions
- The **identity function** \(I_Y\) on \(Y\) is defined by
\[
I_Y(y) \;=\; y \quad \text{for every } y \in Y.
\]
- The composition \(I_Y \circ f\) is the function from \(X\) to \(Y\) given by
\[
(I_Y \circ f)(x) \;=\; I_Y\bigl(f(x)\bigr).
\]
---
## 2. Proof
Take any \(x \in X\). Then
\[
(I_Y \circ f)(x)
\;=\;
I_Y\bigl(f(x)\bigr)
\;=\;
f(x),
\]
because \(I_Y\) just returns its argument unchanged.
Since this holds for every \(x \in X\), the two functions \(I_Y \circ f\) and \(f\) have the same value at every point of \(X\). Therefore,
\[
I_Y \circ f = f.
\]
---
### Conclusion
By definition of composition and the property of the identity map \(I_Y\), we see that composing \(f\) on the right with \(I_Y\) leaves \(f\) unchanged, proving
\[
I_Y \circ f = f.
\]
