Let $F,G$ functors from a category $\mathscr I$ to a category $\mathscr C$ and let $\tau:F\to G$ be a natural trasnformation.
Question: under what conditions the image of $\tau$ can be computed pointwise?
Here an example. Assume that for each object $n$ in $\mathscr I$, the morphism $\tau_n$ has image $m_n:M(n)\to G(n)$ so that $\tau_n=m_n\circ e_n$. If for each $i:n\to k$ in $\mathscr I$, the morphism $G(i):G(n)\to G(k)$ is monic, then there exists a unique morphism $M(i):M(n)\to M(k)$ which makes $M$ a functor and $e:F\to M$, $m:M\to G$ be natural transformations. If the pointwise evaluation preserve monomorphisms, then $m$ is the image of $\tau$.
There are other situations where this happen?
Warning: the terminology below is not standard – I've extracted it myself from the notion of strong epimorphism.
Definition. An image of a morphism $A\xrightarrow{f}C$ is an initial subobject of $C$ among those subobjects of $C$ through which $f$ factors. An image factorization of $A\xrightarrow{f}C$ is a factorization $f=m\circ e$ for $m$ an image of $f$. We say a monomorphism/subobject $m\colon B\hookrightarrow C$ is a strong image of a morphism $A\xrightarrow{f}C$ if $A\xrightarrow{f}C$ factors through $m\colon B\hookrightarrow C$, and if whenever a composite $q\circ f$ factors through some monomorphism/subobject $m'\colon X\to Y$, the composite $q\circ m$ factors as $m'\circ d$ for some (actually unique) $d\colon B\to X$. A factorization $m\circ e$ of $f$ through a strong image of $f$ is called a strong image factorization. $$\require{AMScd} \begin{CD} A @= A @>p>> X\\ @VeVV @VfVV @Vm'VV\\ B@>m>>C@>q>>Y \end{CD} $$ (In the diagram, $B\xrightarrow{m}C$ and $X\xrightarrow{m'}Y$ are monomorphisms and there should be a dashed diagonal $B\xrightarrow{d}X$). Note that $A\xrightarrow{e}B$ does not have to be an epimorphism or even have an image itself.
Lemma. A strong image of a morphism $A\xrightarrow{f}B$ is an initial subobject of $B$ through which $f$ factors. In other words, being strong is a property of the image of $f$.
Proof. The condition of being an initial subobject through which $f$ factors is obtained by restricting the defining property of a strong image to the case where $q=\mathrm{id}_C$.
Theorem. If evaluation preserves monomorphisms (e.g. if the target category has kernel pairs), then strong image factorizations of natural transformations are computed pointwise: if the components $FX\xrightarrow{\phi_X}GX$ of $F\overset\phi\Rightarrow G$ have strong images, then a choice of these strong images $GX\xrightarrow{\mu_X}HX$ of $FX\xrightarrow{\phi_X}GX$ extends naturally to a functor $G$ with a natural transformation $G\overset\mu\Rightarrow H$ that is a strong image of $F\overset\phi\Rightarrow H$.
Sketch of proof of functoriality of $G$. To extend the choices of strong images $GX\xrightarrow{\mu_X}HX$ to a functor, consider the diagram $$\begin{CD} FX_1 @= FX_1 @>Ff>> FX_2 @>>> GX_2\\ @VVV @V\phi_{X_1}VV @V\phi_{X_2}VV @V\mu_{X_2}VV\\ GX_1 @>\mu_{X_1}>> HX_1 @>Hf>> HX_2 @= HX_2 \end{CD}$$ The left and right squares commute by being image factorizations, while the middle square commutes by naturality of $F\overset\phi\Rightarrow H$. Notice that there since $GX_2\xrightarrow{\mu_{X_2}}HX_2$ is a monomorphism, there is at most one diagonal morphism $GX_1\xrightarrow{Gf}GX_2$ that would make the diagram commute. With a little effort, one can also draw the appropriate diagram to see that if such morphisms $GX_1\xrightarrow{Gf}GX_2$ and $GX_2\xrightarrow{Gg}GX_3$ exist, then their composite would be the unique $GX_1\xrightarrow{G(g\circ f)}GX_3$.
Now, since the left square is a strong image factorization, the universal property of strong image factorizations asserts that the unique diagonal morphism $GX_1\xrightarrow{Gf}GX_2$ that makes the diagram commute does indeed exist. Note that this does not require evaluation to preserve monomorphisms.
Sketch of proof of strong image property of $G\overset\mu\Rightarrow H$. Consider the diagram of morphisms $$\begin{CD} FX @= FX @>\rho_X>> JX\\ @V\epsilon_X VV @V\phi VV @V\mu'_X VV\\ GX @>\mu>> H_X @>\kappa_X>> KX \end{CD}$$ where the components of $F\overset\epsilon\Rightarrow G\overset\mu\Rightarrow H$ are strong image factorizations of the components of $F\overset\phi\Rightarrow H$ and $J\overset{\mu'}\Rightarrow K$ is a monomorphism. Since evaluation preserves monomorphisms, $JX\xrightarrow{m'}KX$ are monomorphisms. Hence by the universal property of strong monomorphisms we have unique morphisms $GX\xrightarrow{d_X}JX$ that make the diagrams commute. With a bit of effort, you can draw the appropriate diagram to check these form a natural transformation $G\overset\delta\Rightarrow J$.
Remark. I suspect it is also the case that strong image factorizations are computed pointwise in comma categories $(F\downarrow G)$ when $F$ and $G$ preserve strong images.