TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

First isomorphism theorem for vector spaces: $V / \ker T$ is isomorphic to $im T$

Statement

For every linear map $T : V \to U$ , the formula $\tilde{T} (v + \ker T) : = T (v)$ defines a linear isomorphism $\tilde{T} : V / \ker T \to im T$ .

Facts & Assumptions

Given: A linear map $T : V \to U$ .

[L1]

A linear map whose kernel contains a subspace $W$ factors uniquely through $V / W$ by $v + W \mapsto T (v)$ (Universal property of the quotient vector space).

[L2]

The image of a linear map is a linear subspace, and a linear map is injective exactly when its kernel is the zero subspace (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial).

Proof

technique · direct

1.1L1L2

Apply [L1] with $W = \ker T$ and codomain restricted to $im T$ to obtain the linear map $\tilde{T} (v + \ker T) = T (v)$ ; it is surjective by the definition of $im T$ .

2.1step 1.1L2∎

Its kernel consists of cosets $v + \ker T$ with $T (v) = 0$ , hence only the zero coset $\ker T$ ; [L2] makes $\tilde{T}$ injective, so it is an isomorphism, including the zero map and the zero-space case.

Remarks

For finite-dimensional $V$ , taking dimensions in this isomorphism gives $\dim V = \dim \ker T + \dim im T$ , the equality recorded independently as Rank-nullity: $\dim_{F} V = nullity T + rank T$ . This is an agreement record, not a premise in the proof above.

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Cornell Math 4330, Quotient Spaces, Exercise QuoSpace 5 (standard reference, not scraped)
S. Axler, Linear Algebra Done Right, 4th ed., Results 3.106-3.107 (standard reference, not scraped)

First isomorphism theorem for vector spaces: V/ker⁡TV/kerT is isomorphic to im⁡TimT