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Let $\mathcal C$ be a category. Recall that a morphism $f : X \to Y$ is epi if $$\circ f: \hom(Y,Z) \to \hom(X,Z)$$ is injective for each object $Z \in \mathcal C$. ($f$ is mono if $f\circ : \hom(Z,X) \to \hom(Z,Y)$ is injective.)

Let $\mathcal C,\mathcal D$ be categories. Then $\hom(\mathcal C,\mathcal D)$, the collectional of all functors $\mathcal C \to \mathcal D$, is naturally a category, where the morphisms are natural transformations: if $F,G: \mathcal C \to \mathcal D$ are functors, a natural transformation $\alpha: F \Rightarrow G$ assigns a morphism $\alpha(x) : F(x) \to G(x)$ in $\mathcal D$ for each object $x \in \mathcal C$, and if $f: x \to y$ is a morphism in $\mathcal C$, then $\alpha(y) \circ F(f) = G(f) \circ \alpha(x)$ as morphisms in $\mathcal D$.

Given a natural transformation, can I check whether it is epi (or mono) by checking pointwise? I.e.: is a natural transformation $\alpha$ epi (mono) iff $\alpha(x)$ is epi (mono) for each $x$?

If not, is there an implication in one direction between whether a natural transformation is epi and whether it is pointwise-epi?

A more general question, one that I never really learned, is what types of properties of a functor are "pointwise" in that they hold for the functor if they hold for the functor evaluated at each object. E.g.: is the (co)product of functors the pointwise (co)product?

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up vote 22 down vote accepted

Theo, the answer is basically "yes". It's a qualified "yes", but only very lightly qualified.

Precisely: if a natural transformation between functors $\mathcal{C} \to \mathcal{D}$ is pointwise epi then it's epi. The converse doesn't always hold, but it does if $\mathcal{D}$ has pushouts. Dually, pointwise mono implies mono, and conversely if $\mathcal{D}$ has pullbacks.

The context for this --- and an answer to your more general question --- is the slogan

(Co)limits are computed pointwise.

You have, let's say, two functors $F, G: \mathcal{C} \to \mathcal{D}$, and you want to compute their product in the functor category $\mathcal{D}^\mathcal{C}$. Assuming that $\mathcal{D}$ has products, the product of $F$ and $G$ is computed in the simplest possible way, the 'pointwise' way: the value of the product $F \times G$ at an object $A \in \mathcal{C}$ is simply the product $F(A) \times G(A)$ in $\mathcal{D}$. The same goes for any other shape of limit or colimit.

For a statement of this, see for instance 5.1.5--5.1.8 of these notes. (It's probably in Categories for the Working Mathematician too.) See also sheet 9, question 1 at the page linked to. For the connection between monos and pullbacks (or epis and pushouts), see 4.1.31.

You do have to impose this condition that $\mathcal{D}$ has all (co)limits of the appropriate shape (pushouts in the case of your original question). Kelly came up with some example of an epi in $\mathcal{D}^\mathcal{C}$ that isn't pointwise epi; necessarily, his $\mathcal{D}$ doesn't have all pushouts.

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In all the preceding answers the category $D$ is required to have pushouts / pullbacks in order to epi / mono being equivalent to pointwise epi / mono in functor categories $D^C$.

The discussion in On a corollary in Mitchell's book draw my attention to another important class of categories that usually doesn't have pushouts or pullbacks and where epi / mono is also equivalent to pointwise epi / mono in $D^C$: That is when $D$ is exact.

A category is exact, if

  • it has a zero object
  • kernels and cokernels exist
  • every monomorphism is a kernel and every epimorphism is a cokernel
  • every morphism can be written as a composition of an epimorphism followed by a monomorphism.

As a reference see Barry Mitchell: "Theory of Categories", II.11 Functor Categories.

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I guess you just need some way of using limits to detect monos. Pullbacks give one (via kernel pairs) and every mono being a kernel gives another. – Ben Millwood May 5 '13 at 11:35

The accepted answer is good. If you would like another reference see section 2.15 of the Handbook of categorical algebra, volume I (by F. Borceux) pages 87--90. In particular, their Corollary 2.15.3 tells us the following:

Let $F,G: \mathcal{C} \rightarrow\mathcal{D}$ be two functors where $\mathcal{C}$ is a small category and $\mathcal{D}$ has pullbacks. Then a natural transformation $\alpha : F \rightarrow G$ is monic in $\mathcal{D}^{\mathcal{C}}$ if and only if for each object $C\in \mathcal{C}$, $\alpha_C : F(C)\rightarrow G(C)$ is monic in $\mathcal{D}$.

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This is only a partial answer. Regarding your first question (mono/epi iff pointwise mono/epi): At least for the case where the target category $\mathcal{D}$ is $\mathbf{Set}$, it is true that pointwise mono/epi implies mono/epi, see p. 91 of the 1998 edition of Mac Lane.

As for the second question, the answer is that pointwise limits implies limits in functor categories, by the ``limit with parameters'' theorem (Theorem V.3.1 of Mac Lane).

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After the edit, I've noticed that Tom Leinster also answered the question. So now you have the answer of an expert :). I have decided not to delete my answer, hoping that it may help in some way. – user2734 Mar 12 '10 at 12:18

The answere is "Yes" if the caegory has pullbaks, otherwise no in general: Let $C$ a category and $I$ a diagram, If $C$ ha limits (of some kind) then the some happen for $C^I$ because you can make limits pointwise. Then if $m: F\Rightarrow G$ is Mono (in $C^I$ ) the pullabck of $(m,m)$ give the cospan $F\leftarrow^{1_F}F\to^{1_F}F$ and if $C$ has pullback then you can see what above componentwise, then any pullback of $(m(i), m(i))$ give the cospan $F(i)\leftarrow^{1_{F(i)} }F(i)\to^{1_{F(i)}}F(i)$ then any $m(i)$ is Mono. Now, let $C$ the follow category where no identity arrow are as: $f,g: A\to B$, $h: B\to C$ where $h\circ f= h\circ g $ then in $C^\to$ we have the Monomorphism $(f, h): g\to h$ but $h$ isnt Mono, by the way you get the not pointwise pullback of $(f,h), (f,h))$ give the cospan $g\to^1 g\\leftarrow^{1 } g$ (I take this argument from G Kelly:

Sorry for my poor English, and Latex , but I use xymatrix for make diagram and seems dont work here..

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+1 for the counterexample! – Hew Wolff Aug 13 '15 at 3:24

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