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Identity map

Click a tile: formula, recovery, and proof check appear on the right.

Study route / 学习路线

Fast loop: symbols → theorem → construction H → recovery check → English exam template.

1 Decode the symbols Most mistakes come from confusing a function, a pair, and a subset.
2 Read the full theorem Keep the seven statements separate. Each one transforms a different kind of mathematical object.
3 Click the identity map Ask: input of H, output of H, and recovery map.
4 Use finite counts only as intuition Use counts for intuition only; maps prove the theorem.
5 Finish with recovery tests If you can write the inverse or explain injectivity without looking, the theorem has become usable.

Prerequisite dictionary / 先修词典

Only these terms are needed for the page.

Function space A → B

All functions with input set A and output set B. Each a∈A gets exactly one B-value.

Equinumerous A ≈ B

There exists a bijection A→B: no duplicates and no omissions.

Cardinal comparison A ≤ B

There exists an injection A↪B. B has room for a copy of A.

Cartesian product A × B

The set of ordered pairs (a,b) with a∈A and b∈B. A function A×B→C accepts a pair as one input.

Power set 𝒫(A)

The set of all subsets of A. An element of 𝒫(A) is itself a set X with X⊆A.

Indicator function χX:A→{0,1}

χX(a)=1 when a∈X and χX(a)=0 when a∉X. It stores subset membership as yes/no data.

Well-defined H(x) has one legal value

The formula gives one legal output in the claimed target set.

Empty function ∅ → B

There is exactly one function from ∅ to any set B. If A≠∅ and B=∅, there are no functions A→B.

Theorem 1, uncompressed

Do not memorize seven isolated facts. Each statement preserves the same data in a new shape.

Core slogan same information ⇒ same cardinality

For S≈T, define H:S→T and recover the original object. For S≤T, define H:S→T and show distinct inputs do not collapse.

Relabel both sides A₁≈A₂, B₁≈B₂ ⇒ (A₁→B₁)≈(A₂→B₂)

Bijections on the input and output sets induce a bijection between the function spaces.

Monotone embedding A₁≤A₂, B₁≤B₂ ⇒ (A₁→B₁)≤(A₂→B₂)

This holds when B₂≠∅; if B₂=∅, it fails exactly when A₁=∅ and A₂≠∅.

Currying ((A×B)→C)≈(A→(B→C))

A two-input function can be read as: first choose a, then choose a function of b.

Pairing (A→B×C)≈(A→B)×(A→C)

One function returning pairs is the same as two coordinate functions with common domain A.

Two-point sampling ({0,1}→A)≈A×A

A function on {0,1} is determined by the ordered pair of values (f(0),f(1)).

Power-set transport A≈B ⇒ 𝒫(A)≈𝒫(B)

A bijection f:A→B moves each subset X⊆A to its image f[X]⊆B.

Indicator correspondence (A→{0,1})≈𝒫(A)

A 0/1 function records exactly which elements of A are selected.

Transferable proof spine

Most proofs use the same spine: define H, preserve information, recover the object.

Relabel

Conjugate a function by known bijections.

H(h)=g∘h∘f⁻¹

Changing the names of inputs and outputs does not change how many functions exist.

Curry

Replace a pair input by staged inputs.

H(F)(a)(b)=F(a,b)

The same table of values is read by rows first, then columns; no value is lost.

Split

Turn paired outputs into two functions.

H(F)=(π₁∘F, π₂∘F)

Every A-input produces a B-coordinate and a C-coordinate, which can be stored separately.

Sample

A two-point domain stores an ordered pair.

H(f)=(f(0),f(1))

There are only two inputs, so knowing the two sampled values determines the function.

Transport

Move a subset pointwise through f.

H(X)={f(a) | a∈X}

A bijection on elements automatically induces a bijection on collections of elements.

Indicate

Use 1 exactly on the selected elements.

H(f)={a∈A | f(a)=1}

A 0/1 function and a subset are two notations for the same information.

Construction ledger / 构造账本

Open each row for the forward map, recovery map, and proof check.

Relabel domain and codomain (A₁→B₁)≈(A₂→B₂) bijective
Forward map

H(h)=g∘h∘f⁻¹

Here f:A₁→A₂ and g:B₁→B₂ are bijections.

Recovery

h=g⁻¹∘H(h)∘f

Apply f to move into A₂, then undo g at the output.

Check

f⁻¹ and g⁻¹ exist because f and g are bijections. This is the whole reason the construction is reversible.

Monotone embedding (A₁→B₁)≤(A₂→B₂) injective
Forward map

H(h)(a₂)=g(h(f⁻¹(a₂))) on f[A₁], else b

Use injections f:A₁↪A₂, g:B₁↪B₂ and a fixed b∈B₂.

Recovery test

If H(h)=H(k), compare values on f(a₁). Then g(h(a₁))=g(k(a₁)), so h(a₁)=k(a₁).

Check

The fallback b exists only when B₂≠∅. If B₂=∅, use the edge-case table instead of this extension proof.

Currying ((A×B)→C)≈(A→(B→C)) bijective
Forward map

H(F)(a)(b)=F(a,b)

Turn one pair-input function into a function that returns B→C functions.

Recovery

F(a,b)=G(a)(b)

Given G:A→(B→C), evaluate at a, then at b.

Check

No value changes. Only parentheses in the input process change: (a,b) versus a then b.

Pairing (A→B×C)≈(A→B)×(A→C) bijective
Forward map

H(F)=(π₁∘F,π₂∘F)

Split each pair F(a)=(b,c) into its B and C coordinates.

Recovery

F(a)=(u(a),v(a))

Given coordinate functions u:A→B and v:A→C, pair their values.

Check

The two output coordinates share the same input a. Losing that shared domain would lose the original pairing.

Binary sampling ({0,1}→A)≈A×A bijective
Forward map

H(f)=(f(0),f(1))

Read the two values of f as an ordered pair.

Recovery

f(0)=x, f(1)=y

Given (x,y)∈A×A, define the unique function with those two values.

Check

Order matters. (x,y) and (y,x) can define different functions when x≠y.

Power-set transport A≈B⇒𝒫(A)≈𝒫(B) bijective
Forward map

H(X)={f(a) | a∈X}

Move every selected A-element through the bijection f:A→B.

Recovery

X={f⁻¹(b) | b∈Y}

Move each selected B-element back through f⁻¹.

Check

Injectivity prevents merging two selected points; surjectivity ensures every B-subset can be reached.

Indicator map (A→{0,1})≈𝒫(A) bijective
Forward map

H(f)={a∈A | f(a)=1}

Keep exactly the points where f says yes.

Recovery

χX(a)=1 if a∈X, else 0

Every subset X⊆A has a characteristic function χX.

Check

A 0/1 function is determined pointwise by its 1-set, and every subset has exactly one 1-set function.

Finite sanity check

For finite sets, |A→B|=|B||A|. Use counts as intuition, not proof.

Function space count |A→B| = 2³ = 8

Each element of A independently chooses one element of B.

Currying count 4^(3·2) = (4²)³ = 4096

Choosing a C-value for each pair (a,b) equals choosing a B→C function for each a.

Pairing count (2·4)³ = 2³·4³ = 512

A B×C-output is just two coordinated outputs stored side by side.

Indicator count |A→{0,1}| = 2³ = |𝒫(A)|

Every yes/no membership pattern describes exactly one subset.

What is preserved? The complete table of values of the function, only re-indexed, regrouped, split, or read as membership data.
How to prove it? Write the forward map H and the recovery map. If both recover the original object, H is a bijection.
Where to be careful? For ≤, extending outside f[A₁] needs a default b∈B₂; the empty-codomain caveat is not cosmetic.

Recovery lab

A bijection proof is not complete until the original object can be recovered.

Currying inverse

Given G:A→(B→C), recover F:(A×B)→C.

F(a,b) = ?

Think: G first accepts a, then returns a B→C function.

F(a,b)=G(a)(b). This exactly reverses H(F)(a)(b)=F(a,b).
Pairing inverse

Given (u,v)∈(A→B)×(A→C), recover F:A→B×C.

F(a) = ?

Think: the two coordinate functions share the same input a.

F(a)=(u(a),v(a)). Projection then pairing are inverse operations.
Indicator inverse

Given X⊆A, recover f:A→{0,1}.

f(a) = ?

Think: the subset tells you exactly where the function should be 1.

f(a)=1 if a∈X, and f(a)=0 if a∉X. This is the characteristic function χX.
Injection caveat

Why does monotonicity ask for B₂≠∅?

Outside f[A₁], H(h) must output what?

Think: an extension over A₂ needs a legal value at every new point.

It needs a fixed fallback b∈B₂. If B₂=∅ and A₂ has points outside f[A₁], no legal output exists.

Worked examples / 例题拆解

Use these as short English proof patterns.

Example 1

Show that ({0,1}→A)≈A×A.

  1. Object on the left: a function f with exactly two inputs, 0 and 1.
  2. Forward map: send f to the ordered pair (f(0),f(1)).
  3. Reverse map: given (x,y), define f(0)=x and f(1)=y.
  4. Conclusion: the two procedures undo each other, so this is a bijection.
Example 2

Explain why (A→{0,1}) behaves like 𝒫(A).

  1. A function f:A→{0,1} answers yes/no for each a∈A.
  2. Collect the yes-points: H(f)={a∈A | f(a)=1}.
  3. Given a subset X⊆A, recover f by setting f(a)=1 exactly on X.
  4. Conclusion: choosing a subset is the same as choosing a 0/1-valued function.
Example 3

Use currying without changing any values.

  1. Start with F:(A×B)→C. It accepts one input pair (a,b).
  2. Define G=H(F) by G(a)(b)=F(a,b).
  3. Given G, recover F by F(a,b)=G(a)(b).
  4. Conclusion: currying changes the order of asking for inputs, not the table of C-values.

English exam templates

Use these as compact answer patterns. Replace the set names as needed.

General bijection template

To prove S≈T

  1. Define a map H:S→T.
  2. Define a recovery map R:T→S.
  3. Check R(H(s))=s and H(R(t))=t.
  4. Therefore H is a bijection.
10-second final check

Every answer must close the loop

  1. Name the source and target of H.
  2. Show how to recover the original object.
  3. State the exact conclusion: bijection, injection, or cardinal inequality.
Currying

((A×B)→C)≈(A→(B→C))

  1. Given F:(A×B)→C, define H(F)(a)(b)=F(a,b).
  2. Given G:A→(B→C), recover F by F(a,b)=G(a)(b).
  3. These two constructions undo each other.
Pairing

(A→B×C)≈(A→B)×(A→C)

  1. Map F to (π₁∘F,π₂∘F).
  2. Given (u,v), recover F by F(a)=(u(a),v(a)).
  3. The shared input a preserves the original pairing.
Indicator

(A→{0,1})≈𝒫(A)

  1. Map f to H(f)={a∈A | f(a)=1}.
  2. Given X⊆A, recover f by setting f(a)=1 on X and 0 outside X.
  3. This is exactly the characteristic function of X.
Monotone embedding

(A₁→B₁)≤(A₂→B₂)

  1. Use injections f:A₁↪A₂ and g:B₁↪B₂.
  2. Choose b₀∈B₂. Define H(h)(a₂)=g(h(f⁻¹(a₂))) on f[A₁], and b₀ outside f[A₁].
  3. If H(h)=H(k), compare values on f(a₁); injectivity of g gives h(a₁)=k(a₁).
  4. This construction needs B₂≠∅; otherwise use the edge-case table.

Mastery checks / 掌握度检查

Mastery standard: you can write H, identify its target, and recover the input.

Quick diagnostics

Before revealing answers, say the missing construction aloud.

1. Given G:A→(B→C), what is the original F:(A×B)→C?
2. Given X⊆A, what is the corresponding f:A→{0,1}?
3. In the ≤ theorem, why can’t we ignore the case B₂=∅?
1. F(a,b)=G(a)(b). 2. f(a)=1 on X and 0 outside X. 3. The extension outside f[A₁] needs a default b∈B₂; if no such b exists, the construction may fail.
What you should now be able to do

Completion checklist

  • Explain the difference between A≈B and A≤B.
  • Identify the source and target of H for each theorem item.
  • Write at least one inverse map for currying, pairing, and indicators.
  • Use the finite-count panel as intuition without mistaking it for proof.
  • Handle empty codomains in (A→B) without guessing.

Empty-codomain microscope

Monotonicity usually extends by choosing a fallback b∈B₂. If B₂=∅, split cases directly.

Assuming B₂=∅ Source (A₁→B₁) Target (A₂→B₂) Injection?
A₁=∅ and A₂=∅ one empty function one empty function yes
A₁≠∅ and A₁≤A₂ no functions no functions yes, ∅ embeds into ∅
A₁=∅ and A₂≠∅ one empty function no functions no: singleton cannot inject into ∅

Correct caveat: when B₂=∅, monotonicity fails exactly in the edge case A₁=∅ and A₂≠∅.

Notation rail

These abbreviations compress the theorem into the usual algebra of cardinals and function spaces.

Product square A² := A × A Ordered pairs of A-elements.
Product cube A³ := A × A × A Ordered triples; continue similarly for higher powers.
Binary exponent 2ᴬ := (A → {0,1}) Equivalent to choosing a subset of A.
Function exponent Bᴬ := (A → B) The full function space from A to B.