We prove \((i)\). Consider a pair of morphisms \(u\), \(v\colon Q\to A\) and a 2-cell \(\lambda \colon f\circ u\Longrightarrow f\circ v\). We must show that there is a unique 2-cell \(\xi \colon u\Longrightarrow v\) with \(f\star \xi =\lambda \). Starting from \(\lambda \) we form the following pasting: Since \(h\) is a fully faithful arrow, there is a unique 2-cell \(\varphi \colon u\Longrightarrow v\) such that \(h\star \varphi \) coincides with this pasting. Then \(f\star \varphi =\lambda \): as \(g\) is faithful it suffices to check this after postcomposing with \(g\), and there it holds, by pasting with the given invertible 2-cell \(h\cong g\circ f\). Moreover \(\varphi \) is the unique such 2-cell, since any two 2-cells \(\xi \) with \(f\star \xi =\lambda \) have equal whiskerings with \(h\), and \(h\) is faithful.

We now prove \((ii)\). Say \(h\) is the 2-kernel of \(\ell \colon C\to D\); we prove that \(f\) is the 2-kernel of \(\ell \circ g\). The 2-dimensional universal property is \((i)\), a 2-monomorphism being in particular faithful. For the 1-dimensional one, note first that \(g\circ f\) is a 2-kernel of \(\ell \), by Corollary 2.22; in particular \(\ell \circ g\circ f\) is isomorphic to a null morphism. Let \(r\colon M\to B\) be such that \((\ell \circ g)\circ r\) is isomorphic to a null morphism. Since \(g\circ f\) is the 2-kernel of \(\ell \), there are \(s\colon M\to A\) and an invertible 2-cell \(\beta \colon g\circ f\circ s\cong g\circ r\); as \(g\) is fully faithful, \(\beta \) lifts to an invertible 2-cell \(f\circ s\cong r\). So \(f\) satisfies the universal property of the 2-kernel of \(\ell \circ g\).

Let \(k\colon K\to A\) be a 2-kernel of \(f\colon A\to B\), let \(u\colon K'\to K\) and \(w\colon A\to A'\) be equivalences and let \(w'\) be a quasi-inverse of \(w\). We claim that \(w\circ k\circ u\) is a 2-kernel of \(f\circ w'\). It is a 2-monomorphism, being a composite of 2-monomorphisms, so condition \((2)\) of Definition 2.18 holds; and \(f\circ w'\circ w\circ k\circ u\cong f\circ k\circ u\) is null. If \(z\colon Z\to A'\) satisfies \(f\circ w'\circ z\cong 0\), then the universal property of \(k\) yields \(s\colon Z\to K\) with \(w'\circ z\cong k\circ s\), whence \(z\cong w\circ k\circ s\cong (w\circ k\circ u)\circ (u'\circ s)\) for a quasi-inverse \(u'\) of \(u\). This proves the first assertion, since every equivalence is invertible up to isomorphism on either side; the second is dual.

For the consequence, let \(w\cong m\circ e\) be normal. Then \(v\circ w\circ u\cong (v\circ m)\circ (e\circ u)\) is a normal 2-monomorphism after a normal 2-epimorphism, hence normal; and the converse follows by composing with quasi-inverses. The same computation applies with the factors interchanged, which is antinormality.

Say \(f\colon A\to B\), \(g\colon A\to Q\) and \(m\colon Q\to B\). Consider the 2-kernel \(K\xrightarrow []{k}A\) of \(g\). Certainly \(f\circ k\) is isomorphic to a null morphism, since \(g\circ k\) is. Let \(r\colon M\to A\) be a morphism such that \(f\circ r\) is isomorphic to a null morphism. Then so is \(m\circ g\circ r\), and \(m\) reflects null morphisms by Proposition 2.16, so \(g\circ r\) is isomorphic to a null morphism as well. The universal property of the 2-kernel of \(g\) now yields \(s\colon M\to K\) and an invertible 2-cell \(r\cong k\circ s\). Since \(k=2\text{-}\mathrm{ker}(g)\) is a 2-monomorphism, this exhibits \(k\) as a 2-kernel of \(f\).

Conversely, consider the 2-kernel \(H\xrightarrow []{h}A\) of \(f\). Since \(f\circ h\) is isomorphic to a null morphism, so is \(m\circ g\circ h\), and hence so is \(g\circ h\), again by Proposition 2.16. That \(h\) is a 2-kernel of \(g\) now follows in the same way, the 2-dimensional universal property being automatic since \(h\) is a 2-kernel of \(f\). The second part of the statement is dual.

We prove the first statement; the second follows by duality. Let \(q\colon B\to Q\) be a 2-cokernel of a morphism \(f\colon A\to B\), with invertible 2-cell \(\eta \colon q\circ f\cong 0\), and let \((k\colon K\to B,\ \kappa \colon q\circ k\cong 0)\) be the 2-kernel of \(q\). We verify that \((q,\kappa )\) is a 2-cokernel of \(k\).

Condition (1) of Definition 2.18, applied to the datum \((f,\eta )\), provides a morphism \(w\colon A\to K\) and an invertible 2-cell \(\gamma \colon k\circ w\cong f\). Let now \(z\colon B\to Z\) be a morphism with an invertible 2-cell \(\beta \colon z\circ k\cong 0\). Then \((\beta \star w)\cdot (z\star \gamma ^{-1})\colon z\circ f\cong 0\) is invertible, so condition (1) of Definition 2.19 applied to \(q\) yields a morphism \(u\colon Q\to Z\) and an invertible 2-cell \(u\circ q\cong z\). This is condition (1) for the 2-cokernel of \(k\). Condition (2) for the 2-cokernel of \(k\) is literally condition (2) of \(q=2\text{-}\mathrm{coker}(f)\). Hence \(q\) is a 2-cokernel of \(k\).

We prove the first equivalence; the second is dual.

Suppose \(K\) is trivial, so that \(\operatorname {id}_{K}\cong 0\) and hence \(k=k\circ \operatorname {id}_{K}\) is isomorphic to a null morphism. Then for every \(z\colon A\to Z\) the composite \(z\circ k\) is isomorphic to a null morphism, so condition (1) of Definition 2.19 yields \(u\colon Q\to Z\) with \(u\circ q\cong z\); this says that \((-)\circ q\colon {\mathit{L}}\left({Q},\, {Z}\right)\to {\mathit{L}}\left({A},\, {Z}\right)\) is essentially surjective. Condition (2) says that it is fully faithful. As this holds for every \(Z\), Lemma 2.15 makes \(q\) an equivalence.

Conversely, suppose \(q\) is an equivalence, with quasi-inverse \(p\). Whiskering \(\kappa \colon q\circ k\cong 0\) with \(p\) gives \(p\circ q\circ k\cong 0\), whence \(k\) is isomorphic to a null morphism. Now \(k\) is a 2-kernel, so a 2-monomorphism by Proposition 2.21, and therefore reflects null morphisms; applied to \(k\circ \operatorname {id}_{K}=k\) this gives \(\operatorname {id}_{K}\cong 0\), that is, \(K\) is trivial.

We prove the first statement; the second is dual. By Proposition 3.9 we have \(2\text{-}\mathrm{ker}(f)\simeq 2\text{-}\mathrm{ker}(e)\), so the 2-kernel of \(e\) is trivial. As \(e\) is a normal 2-epimorphism it is the 2-cokernel of its 2-kernel, by Proposition 3.11, so that \(2\text{-}\mathrm{ker}(e)\) and \(e\) form a short 2-exact sequence. Proposition 3.15 applied to that sequence shows \(e\) to be an equivalence, whence \(f\cong m\) is a normal 2-monomorphism.

Let \(q\colon A\to Q\) be a 2-cokernel of \(g\colon G\to A\), with \(\eta \colon q\circ g\cong 0\), and suppose \(q\) is a 2-monomorphism. Then \(q\) reflects null morphisms by Proposition 2.16, so \(g\) is isomorphic to a null morphism. Hence \(\operatorname {id}_{A}\circ g\) is isomorphic to a null morphism, and condition (1) of Definition 2.19 yields \(u\colon Q\to A\) with \(u\circ q\cong \operatorname {id}_{A}\). Whiskering with \(q\) gives \(q\circ u\circ q\cong q\cong \operatorname {id}_{Q}\circ q\), and \(q\), being a 2-cokernel, is a 2-epimorphism by Proposition 2.21; so \(q\circ u\cong \operatorname {id}_{Q}\) by the dual of Remark 2.14. Thus \(q\) is an equivalence. The second statement is dual.

We prove \((i)\Rightarrow (ii)\). The identity \(\operatorname {id}_{B}\) is a 2-kernel of any chosen null morphism \(t\colon B\to 0\): the composite \(t\circ \operatorname {id}_{B}\) is null, every morphism factors through the identity, and \(\operatorname {id}_{B}\) is fully faithful. By Corollary 2.22 the equivalence \(f\) is therefore a 2-kernel of \(t\) as well, and dually a 2-cokernel of any chosen null morphism \(0\to A\). Clearly \((ii)\) implies both \((iii)\) and \((iv)\), and Proposition 3.17 shows that either of these implies \((i)\).

We prove the first part; the second is dual. A normal 2-monomorphism is a 2-monomorphism, and it is normal, being isomorphic to itself composed with the identity. Conversely, if \(f\) is a 2-monomorphism then its 2-kernel is trivial by Lemma 2.24, so Corollary 3.16 makes it a normal 2-monomorphism.

An equivalence is both a 2-monomorphism and a 2-epimorphism, so its 2-kernel and 2-cokernel are trivial by Lemma 2.24. Conversely, let \(f\cong m\circ e\) be normal with trivial 2-kernel and trivial 2-cokernel. By Corollary 3.16, triviality of the 2-kernel makes \(e\) an equivalence and triviality of the 2-cokernel makes \(m\) an equivalence; hence so is \(f\).

Suppose \(f\cong m\circ e\) and \(f\cong m'\circ e'\) as stated. Consider the 2-kernel \(K\xrightarrow []{k}A\) of \(f\), which is a 2-kernel of \(e\) by Proposition 3.9; so \(e\) is a 2-cokernel of \(k\), by Proposition 3.11. Since \(m'\circ e'\circ k\) is isomorphic to a null morphism and \(m'\) is a 2-monomorphism, Proposition 2.16 gives that \(e'\circ k\) is isomorphic to a null morphism too. The universal property of the 2-cokernel \(e\) of \(k\) now yields a morphism \(t\colon I\to I'\) and an invertible 2-cell \(\gamma \colon e'\cong t\circ e\). Moreover, since \(e\) is cofully faithful, the pasting in which the invertible 2-cell on the right is the composite \(m'\circ e'\cong f\cong m\circ e\), induces an invertible 2-cell \(\sigma \colon m'\circ t\cong m\). The situation is this: By part (ii) of Proposition 3.6, since \(m\) is a normal 2-monomorphism and \(m'\) is a 2-monomorphism, \(t\) is a normal 2-monomorphism; and by the dual of part (i), since \(e\) and \(e'\) are 2-epimorphisms, \(t\) is a 2-epimorphism. So \(t\) is an equivalence by Corollary 3.18, and the two factorisations agree up to equivalence.

In the course of this argument we assumed of \(m'\) and \(e'\) only that they are a 2-monomorphism and a 2-epimorphism. Since \(t\) is an equivalence, Corollary 2.22 makes \(m'\) a normal 2-monomorphism and \(e'\) a normal 2-epimorphism.

Let \(m\circ u\cong n\circ p\) be a normal image factorisation. By Proposition 3.9, applied to \(m\circ u\) with the 2-monomorphism \(m\) and to \(n\circ p\) with the 2-monomorphism \(n\), the morphisms \(u\), \(m\circ u\) and \(p\) all have the same 2-kernel \(k\); and \(p\), being a normal 2-epimorphism, is a 2-cokernel of \(k\), by Proposition 3.11. Since \(u\circ k\cong 0\), the universal property of \(p\) yields \(u'\) with \(u'\circ p\cong u\). Then \(m\circ u'\circ p\cong m\circ u\cong n\circ p\), and \(p\) is a 2-epimorphism, so \(m\circ u'\cong n\); as \(n\) is a normal 2-monomorphism and \(m\) is a 2-monomorphism, part (ii) of Proposition 3.6 makes \(u'\) a normal 2-monomorphism. Hence \(u\cong u'\circ p\) is a normal 2-epimorphism followed by a normal 2-monomorphism, so normal.

Whiskering an invertible 2-cell with a morphism yields an invertible 2-cell, and a vertical composite of invertible 2-cells is invertible. Both sides of 3 are therefore invertible 2-cells \(q\circ g\circ k'\Longrightarrow 0\) with the same domain and the same null codomain. By Lemma 2.6 they coincide.

We prove (1). Write \(n\colon N\to A'\) for \(2\text{-}\mathrm{ker}(g)\), so that \(g\circ n\cong 0\). Postcomposing with \(q\) and using \(\varphi _Q\colon h\circ q'\cong q\circ g\) gives \(h\circ q'\circ n\cong 0\); as \(h\) is a 2-monomorphism it reflects null morphisms, so \(q'\circ n\cong 0\). Since \(k'=2\text{-}\mathrm{ker}(q')\), the morphism \(n\) factors as \(n\cong k'\circ m\) for some \(m\colon N\to K'\). Precomposing \(\varphi _K\colon g\circ k'\cong k\circ f\) with \(m\) now gives

so that \(f\circ m\cong 0\) because \(k\) is a 2-monomorphism, and then \(m\cong 0\) because \(f\) is one. Hence \(n\cong k'\circ m\cong 0\), and \(n\), being a 2-kernel and so itself a 2-monomorphism, reflects this to \(\operatorname {id}_{N}\cong 0\). Thus \(2\text{-}\mathrm{Ker}(g)\) is trivial, and since \(g\) is normal, Corollary 3.16 makes it a normal 2-monomorphism.

For (2), write \(p\colon A\to P\) for \(2\text{-}\mathrm{coker}(g)\), so that \(p\circ g\cong 0\). Precomposing with \(k'\) and using \(\varphi _K\) gives \(p\circ k\circ f\cong 0\); as \(f\) is a 2-epimorphism it coreflects null morphisms, so \(p\circ k\cong 0\). Since \(q=2\text{-}\mathrm{coker}(k)\), the morphism \(p\) factors as \(p\cong u\circ q\) for some \(u\colon Q\to P\). Postcomposing \(\varphi _Q\) with \(u\) gives

so that \(u\circ h\cong 0\) because \(q'\) is a 2-epimorphism, and then \(u\cong 0\) because \(h\) is one. Hence \(p\cong u\circ q\cong 0\) and \(\operatorname {id}_{P}\cong 0\), so \(2\text{-}\mathrm{Cok}(g)\) is trivial and \(g\) is a normal 2-epimorphism.

Statement (3) follows from (1) and (2): an equivalence is both a 2-monomorphism and a 2-epimorphism, so \(g\) is at once a normal 2-monomorphism and a normal 2-epimorphism, hence an equivalence by Corollary 3.18.