Inverse Function

Context: FIT1058_MOC · reverses a function · exists only for a bijection · undoes/redoes via Function Composition

Quick Revision

• 🎯 Objective: run a function backwards: given $y$, return the $x$ that produced it ➔ $f^{-1}(y)=$ the unique preimage.
• 📦 Core Components: swap graph pairs ➔ $f^{-1}\circ f=i_A$, $f\circ f^{-1}=i_B$.
• ⚡ Key Constraint: exists iff $f$ is a bijection (injective → uniqueness, surjective → existence).

📝 Core

1. The Inverse (Reverse the Map)

• Definition ➔ $f^{-1}(y)$ = the unique $x$ with $f(x)=y$.
• Requires ➔ every $y$ has exactly one preimage ⟹ $f$ a bijection.
• Graph ➔ swap each pair: $f^{-1}=\{(y,x):(x,y)\in f\}$.

2. Cancellation

• Both ways ➔ $f^{-1}(f(x))=x$, $f(f^{-1}(y))=y$.
• Composition ➔ $f^{-1}\circ f=i_A$, $f\circ f^{-1}=i_B$.

3. Why Bijectivity

• Injective ➔ removes non-uniqueness (several inputs share an output).
• Surjective ➔ removes non-existence (some value never produced).
• Both ➔ each $y$ has one and only one preimage.

Key identities:

$f^{-1}=\{(y,x):(x,y)\in f\},f^{-1}\circ f=i_A,f\circ f^{-1}=i_B$ $(g\circ f{)}^{-1}=f^{-1}\circ g^{-1}(\text{socks-and-shoes})$

⚖️ Core Decision Matrix

Obstacle	Blocked by	Ensures
several inputs share output	injectivity	uniqueness
some value never produced	surjectivity	existence
both removed	bijectivity	$f^{-1}$ is a function
non-bijective $f$	—	only an inverse relation

When It Flips: encryption must be bijective so decryption $=f^{-1}$ exists (Cryptosystem); $x\mapsto x^2$ on $\mathbb{R}$ has no inverse (not injective, not surjective). $f^{-1}$ is itself a bijection $B\to A$.

📊 Exam Execution Trace

Manual Execution Trace

$f=\{(1,b),(2,c),(3,a)\}$:

Step / State	Pair of $f$	Swapped ($f^{-1}$)	Check
0 (Init)	—	—	—
1	$(1,b)$	$(b,1)$	$f^{-1}(f(1))=1$
2	$(2,c)$	$(c,2)$	$f^{-1}(f(2))=2$
3	$(3,a)$	$(a,3)$	$f^{-1}(f(3))=3$

⚠️ Common Mistakes

• 💡 Inverse relation always exists ➔ swapping pairs works for any relation; it’s a function only if $f$ was bijective (else not single-valued or not total).

🧠 Active Recall

Exactly when does $f:A\to B$ have an inverse function, and why each condition?

• Hint: Bijection = uniqueness + existence.

Answer

• Short answer: Iff $f$ is a bijection; injectivity gives a unique preimage, surjectivity guarantees one exists.
• Why: Exactly one preimage ➔ makes $f^{-1}(y)$ well-defined.

How is $f^{-1}$ obtained from the graph, and what equations show it undoes $f$?

• Hint: Swap pairs.

Answer

• Short answer: $f^{-1}=\{(y,x):(x,y)\in f\}$; $f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$.
• Why: Composition ➔ $f^{-1}\circ f=i_A$, $f\circ f^{-1}=i_B$.

A function is not bijective — does it still have an inverse relation?

• Hint: Relation vs function.

Answer

• Short answer: Yes — swapping pairs always yields an inverse relation, just not necessarily a function.
• Why: Fails single-valued/total ➔ non-injective ⟹ multiple preimages; non-surjective ⟹ missing preimages.