Image of commutative diagram is commutative under functor? 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.
 A: 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!
A: 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
A: Some explanations why the claim holds were already given. To show what is wrong with you counterexample:
The "image" of this diagram

is this diagram

and not this one

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.
