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A commuting square $vf=f^{\prime }u$ induces a unique map between their images by $I(u,v)p=p^{\prime }u$ and $i^{\prime }I(u,v)=vi$. The cokernel in a category property supplies existence, and cancellation of the epimorphic $p$ supplies uniqueness. These equations prove identity and composition laws, giving a functor from the arrow category to the abelian category.

In an abelian category, the image factorization in an abelian category of $f:A\to B$ iswhere the abelian-category axiom identifies coimage with image. Thus $p$ is epic and $i$ is monic. Any other epi-mono factorization $f=jq$ has $\ker q=\ker f$ and $\mathrm{coker}j=\mathrm{coker}f$. Since epimorphisms are cokernels of their kernels, its middle object is canonically isomorphic to $\mathrm{im}f$, uniquely compatibly with the two factors.

For a square $vf=f^{\prime }u$, define $I(u,v):\mathrm{im}f\to \mathrm{im}{f}^{\prime }$ byThe first arrow exists because $u\ker f$ factors through $\ker f^{\prime }$, so $p^{\prime }u$ annihilates $\ker f$. Its composite with $i^{\prime }$ equals $vi$ after the epimorphism $p$, proving the second equation. Uniqueness after $p$ proves preservation of identities and composition. This gives the functoriality of abelian image factorization as a functor from the arrow category.

For pullback stability of abelian image factorization, state the standard facts that pullbacks preserve monomorphisms, epimorphisms in an abelian category are stable under pullback, and two adjoining pullback squares have pullback outer rectangle. In the given diagram, $i^{\prime }$ is therefore monic and $p^{\prime }$ is epic, while the composite $i^{\prime }{p}^{\prime }$ is the pullback of $f$. Its epi-mono factorization is an image factorization by the uniqueness just proved. Thus the top row is the image factorization of the pulled-back arrow, with its middle object canonically the pullback of the original image subobject.

Let $B$ be a subterminal object in a Cartesian closed category. For every $X$ and $A$, the exponential object adjunction givesThe set on the right has at most one element because $B$ is subterminal. Hence $B^A$ is subterminal, proving that $\mathrm{Sub}(1)$ is an exponential ideal.

Now let $\mathcal{D}\subseteq \mathcal{C}$ be reflective, with reflector $L$ and unit ${\eta }_A:A\to LA$. Recall the useful form of its universal property: an object $D$ lies in $\mathcal{D}$ precisely when every map $f:A\to D$ factors uniquely through ${\eta }_A$.

Suppose first that $\mathcal{D}$ is an exponential ideal. Reflective subcategories are closed under ambient limits, so $LA\times LB$ lies in $\mathcal{D}$. Given $f:A\times B\to D$ with $D\in \mathcal{D}$, curry in the first variable to obtain $A\to D^B$. Since $D^B\in \mathcal{D}$, this factors uniquely through ${\eta }_A$ and uncurries to $LA\times B\to D$. Curry once more, now in $B$; since $D^{LA}\in \mathcal{D}$, the result factors uniquely through ${\eta }_B$. Thus every $f$ factors uniquely throughThis makes $LA\times LB$ a reflection of $A\times B$, so uniqueness of reflections gives

Conversely, suppose $L$ preserves binary products, and take $D\in \mathcal{D}$. To prove $D^A\in \mathcal{D}$, start with $f:X\to D^A$ and let $\tilde{f}:X\times A\to D$ be its transpose. Since $D$ is reflective, $\tilde{f}$ factors uniquely throughComposing the resulting map $LX\times LA\to D$ with $1_{LX}\times {\eta }_A$ and currying produces an extension $LX\to D^A$ of $f$. Product preservation identifies the unit on $LX\times A$ with $1_{LX}\times {\eta }_A$, so the same universal property proves uniqueness. Therefore $D^A$ is reflective, and $\mathcal{D}$ is an exponential ideal. This proves the reflector product criterion for an exponential ideal.

Finally consider the arrow category $[2,\mathbf{Set}]$. For arrows $a:A_0\to A_1$ and $b:B_0\to B_1$, letThen the exponential $b^a$ is the arrowIndeed, a commutative square from $c:C_0\to C_1$ to this arrow is, after currying, exactly a commutative square $a\times c\to b$. This establishes the required exponential adjunction and proves that $[2,\mathbf{Set}]$ is cartesian closed.

If $b$ is injective, two elements $(u_0,u_1)$ and $(u_0^{\prime },u_1)$ of $F$ haveso injectivity gives $u_0=u_0^{\prime }$. Hence $F\to B_1^{A_1}$ is injective. The category of injective functions is therefore an exponential ideal in the arrow category. Its terminal object and binary products are inherited pointwise, so these same exponential objects make $\mathrm{Mono}(\mathbf{Set})$ cartesian closed.