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.
Best Answer
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!