Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The Wikipedia article for Functor ( http://en.wikipedia.org/wiki/Functor ) claims:

Two important consequences of the functor axioms are (where $F \colon C \to D$ is a covariant functor between categories $C$ and $D$)

  • F transforms each commutative diagram in C into a commutative diagram in D;
  • if f is an isomorphism in C, then F(f) is an isomorphism in D.

The second is obvious. The first one seems plausible, but I can't seem to prove it.

QUESTION 1: Is the first claim even true?

It seems like the following is a counterexample, but I could be missing something. The general construction is a functor that is non-injective on objects and introduces nontrivial homology (thinking of objects and morphisms as 0 and 1-cells in a CW-complex).

Define categories $C$ and $D$ by $$\mathrm{Ob}(C) := \{a_0,b_0,c_0,d_0,a_1,b_1,c_1,d_1\},$$ $$\mathrm{Ob}(D) := \{a,b,c,d\},$$ $$\mathrm{Mor}(C) := \{f\colon a_0 \to b_0, g\colon c_0 \to d_0, x\colon a_1\to c_1, y\colon b_1\to d_1\},$$ $$\mathrm{Mor}(D) := \{\phi\colon a\to b, \psi\colon b\to d, \theta\colon a\to c, \omega\colon c\to d, \mu,\nu\colon a\to d\},$$ where $\psi \circ \phi := \mu$ and $\omega \circ \theta := \nu$, and of course the identity morphisms in each category are understood to exist.

Define a functor $F\colon C \to D$ by $F(a_i) := a$, $F(b_i) := b$, $F(c_i) := c$, $F(d_i) := d$, $F(f) := \phi$, $F(g) := \omega$, $F(x) := \theta$, and $F(y) := \psi$. Again, the functor is understood to take identity morphisms to identity morphisms.

Using the entire category $C$ as the commutative diagram, the image is the category $D$ without the morphisms $\mu$ or $\nu$, and is certainly not a commutative diagram (because the two different paths from $a$ to $d$ render the two different morphisms $\mu$ and $\nu$.

QUESTION 2: Is there an error in this construction?

If the original claim is not true in general, it seems like adding the requirement that the functor be injective on objects would be sufficient.

share|improve this question
3  
Commutative diagram can be viewed as a functor, see Wikipedia. With this viewpoint in mind, the first claim is only the claim that we are able to compose functors. –  Martin Sleziak Aug 6 '12 at 6:46
    
@Martin that's quite succint 8) –  Alexei Averchenko Aug 6 '12 at 6:53

3 Answers 3

up vote 3 down vote accepted

Some explanations why the claim holds were already given. To show what is wrong with you counterexample:

The "image" of this diagram

enter image description here

is this diagram

enter image description here

and not this one

enter image description here

In the other words, that result claims nothing about the diagram in the last picture.

The claim says that whenever something commutes in the first category, the image must commute in the second category. E.g. $f\circ id_{a_0}=f$, hence also $F(f)\circ F(id_{a_0})=F(f)$.

To be able to say something about $F(y)\circ F(f)$ you would need some condition about $y\circ f$ in $C$; such condition cannot be given where since these two morphisms cannot be composed.

share|improve this answer
    
I think that's not quite right -- the endpoints of the arrows for $f,g,x,y,\phi,\psi,\omega,\theta$ are backwards. I think I see what you're getting at by showing two copies of each of $a,b,c,d$ is that the image of a diagram under a functor is distinct from a diagram in the codomain of the functor. If this is the case, then what is the distinction exactly? Using some sort of pullback category? –  Victor Dods Aug 6 '12 at 7:12
    
I exchanged $a_0$ and $a$, $a_1$ and $a$ by mistake - that's what you meant, probably. I've uploaded new versions of the first two diagrams. –  Martin Sleziak Aug 6 '12 at 7:18
    
Yes, that's right. Why though are there two copies of each of $a,b,c,d$ in the second picture? I assume the intent is to imply that certain arrows aren't to be composed when talking about the commutativity of the diagram. But if this is the case, then the notion of commutativity in this case seems too weak to be useful. –  Victor Dods Aug 6 '12 at 7:37
    
Yes, there are two copies on purpose, to stress which arrows should be composed. I still think that this claim about images of commutative diagrams can be useful, but perhaps the most important thing at the moment is to understand the claim correctly. –  Martin Sleziak Aug 6 '12 at 7:41
1  
Ah, ok. In my example, there were no nontrivial commutativity conditions in the domain category, and therefore none were required in the image. Thanks. –  Victor Dods Aug 6 '12 at 7:52

Because of lack of time just an answer to question 1.

Well, it is true. Let $F$ be the functor, and for simplicity take a square commutative diagram. So we have maps $f,g,h,r$ such that $fg=hr$. Then \begin{align*} F(f)F(g) &= F(fg)\\ &= F(hr)\\ &= F(h)F(r) \end{align*} as desired.

The same trick works for any commutative diagram of course; the idea is to use the functor properties to "put all the maps between the brackets" of the functor, then use the relation given by the diagram in the original category, and finally pull the new maps out of the brackets again. This gives us the same relation on maps as in the original category.

Note that i don't really use a drawing here, in fact the "old fashioned" way of just writing an equality of composition of maps is easier here.

Hope that helps, Joachim

share|improve this answer
    
While it's certainly true that if you have a commuting square diagram in the domain category then its image is a commuting square diagram in the codomain category, what happens when there is no such square (or even triangle) in the domain category? In my example, the composition $F(y) \circ F(f)$ is well-defined, but not the composition $y \circ f$ (since the codomain of $f$ is not equal to the domain of $y$). –  Victor Dods Aug 6 '12 at 6:57
    
As i mentioned you don't need a square or triangle, the diagram can be as weird as you want. Just pull the maps into the functor brackets, apply the relation and pull them out again. I looked into your example now and i think you derived a commuting diagram in the target category that is not the image of a diagram in the original category. That is not a contradiction, the statement is just about images of diagrams in the original category. –  Joachim Aug 6 '12 at 7:08
    
The square part consisting of the arrows $\phi, \psi, \theta, \omega$ is a diagram in $D$ which is the image of $C$ under $F$. Or does this fail to be a diagram at all, because it is not closed under composition of arrows? –  Victor Dods Aug 6 '12 at 7:22
    
Reading your comment at Martins answer, i see you understood where the problem is. So i assume i dont need to answer your comment at my question anymore. –  Joachim Aug 6 '12 at 8:00

It is a direct consequence of the fact that functors preserve composition. Preservation of commutative diagrams means just that $$F(a_1) \ldots F(a_n) = F(a_1\ldots a_n) = F(b_1 \ldots b_m) = F(b_1) \ldots F(b_m)$$ for any two paths $a_1 \ldots a_n$ and $b_1 \ldots b_m$ that start and end at the same object.

BTW, did you know Curiosity rover just landed? Why are you doing math right now? Go see the videos!

share|improve this answer
    
Oops, you actually posted your answer less than a minute before i posted mine, sorry for the duplicate.. –  Joachim Aug 6 '12 at 6:50
    
It's ok :D ${}{}$ –  Alexei Averchenko Aug 6 '12 at 6:50
    
+0.25 for correct answer and +0.75 for Curiosity reference :) –  Neal Aug 6 '12 at 7:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.