ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A surjective circle loop can have degree zero and be nullhomotopic

Example

Define $u : I \to R$ by

$u (t) = {\begin{cases} 2 t, & 0 \leq t \leq 1 / 2, \\ 2 - 2 t, & 1 / 2 \leq t \leq 1, \end{cases}$

and put $α = p \circ u$ . The loop $α$ is surjective and nonconstant, but it has degree zero and is nullhomotopic.

Facts & Assumptions

Given: The displayed out-and-back function $u$ and its projection $α = p \circ u$ .

[L1]

The continuous quotient projection satisfies $p (x) = p (y)$ exactly when $x - y \in Z$ , and $p^{- 1} ([0]) = Z$ (The circle as $S^{1} = R / Z$ with basepoint $[0]$ ).

[L2]

Degree is the terminal value of the unique lift beginning at zero (The degree of a based circle loop).

[L3]

A based circle loop is nullhomotopic exactly when its degree is zero (A based circle loop is nullhomotopic exactly when its degree is zero).

[L4]

Functions continuous on each member of a finite closed cover, and agreeing where the pieces meet, paste to a continuous function (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L5]

Constant functions, the identity, finite sums, and scalar multiples are continuous on real intervals (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function).

[L6]

Every real $x$ has a unique integer $m$ with $m \leq x < m + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[L7]

A composite of continuous maps is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

Verification

technique · direct

1.1L4L5algebra

At $t = 1 / 2$ both formulas give $1$ . Each piece is continuous by [L5], so [L4] makes $u$ continuous. Its values at the boundary are $u (0) = 0$ , $u (1 / 2) = 1$ , and $u (1) = 0$ .

2.1step 1.1L1L2L7

By [L1] and [L7], $α = p \circ u$ is a continuous based loop at $[0]$ . The function $u$ is a lift of $α$ beginning at zero, so [L2] gives $\deg (α) = u (1) = 0$ .

3.1step 1.1step 2.1L1L6algebra

Let $[x] \in R / Z$ . By [L6], $r = x - ⌊ x ⌋ \in [0, 1)$ and $p (r) = p (x)$ . For $t = r / 2 \in [0, 1 / 2)$ , the first formula gives $u (t) = r$ , so $α (t) = [x]$ ; hence $α$ is surjective. It is nonconstant because $α (0) = [0]$ while $α (1 / 4) = [1 / 2] \neq [0]$ , the latter inequality following from $1 / 2 \notin Z$ in [L1].

4.1step 2.1step 3.1L3∎

Step 2.1 gives degree zero, so the reverse direction of [L3] makes $α$ nullhomotopic. Step 3.1 shows that nullhomotopy here neither forces constancy nor prevents surjectivity.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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