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As I am very new to category theory, please let me know if I misspeak.

Let C and D be categories, F a functor from C to D, and f a morphism in Hom$_C$(X, Y). From what I can understand, the functor axioms (pasted below) imply that F(f) lives in Hom$_D$(F(X), F(Y)).

i) F(g$\circ$f) = F(g)$\circ$F(f) for all morphisms f : X $\rightarrow$ Y and g : Y $\rightarrow$ Z in Map(C).

ii) F(id$_X$) = id$_{F(X)}$ for all objects, X, in Ob(C).

So far, where F(f) must live, is the only fact I know about functors. I was wondering whether or not there are examples of two distinct objects being mapped (by a functor) to a single object. I tried to use the axioms to get some sort of resolution, but made no progress.

If anyone has any interesting facts about functors, or knows of an example to what was outlined above, I would greatly appreciate it!

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  • $\begingroup$ Map your favorite category with more than one object to the category consisting of one object and its identity morphism. $\endgroup$ – John Douma Aug 15 '18 at 4:00
  • $\begingroup$ That one works for me, thanks $\endgroup$ – Coreymonsta Aug 15 '18 at 4:20
  • $\begingroup$ Less "artificially" there's the functor from groups to sets that takes a group to its underlying set (it just "forgets" the group operations). Since you can end up with many group structures on the same set, you'll generally have a great many distinct groups sent to the same set by this functor. $\endgroup$ – Malice Vidrine Aug 15 '18 at 15:19
  • $\begingroup$ the homomorphisms in Grp could be send to their point-wise equivalent function in Set? $\endgroup$ – Coreymonsta Aug 15 '18 at 20:14
  • $\begingroup$ I was wondering if that functor would be surjective or not out of curiosity, too $\endgroup$ – Coreymonsta Aug 15 '18 at 20:33
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I would comment but I do not yet have enough reputation, so I decided to develop my point a bit more:

Apart from what has been said in the comments about functors that map different objects to the same object, you asked about implications of the functor axioms. Now one certainly can derive that $F(f)$ must be an arrow $F(a) \rightarrow F(b)$ for $f:a \rightarrow b$, however, this is not the whole point of them. Instead, they serve to ensure that functors preserve structure in a certain sense.

Indeed, $F(g \circ f) = F(g) \circ F(f)$ ensures that $F$ preserves commuting triangles: \begin{array}{ccc} a & \xrightarrow{f} & b \\ & \searrow & \downarrow{\scriptstyle g} \\ & & c \end{array} is mapped to \begin{array}{ccc} F(a) & \xrightarrow{F(f)} & F(b) \\ & \searrow & \downarrow{\scriptstyle F(g)} \\ & & F(c) \end{array}

so if the first diagram commutes, it would be necessary for $g \circ f$ to be mapped to $F(g) \circ F(f)$ to ensure that the second diagram also commutes.

Now $\newcommand{\id}{\text{id}} F(\id_a) = \id_{F(a)}$ ensures that $F$ preserves isomorphisms: If $f: a \rightarrow b$ and $g: b \rightarrow a$ are inverses, i.e. $f \circ g = \text{id}_b$ and $g \circ f = \text{id}_a$, then we also want that $F(f)$ be inverse to $F(g)$, i.e. $F(g) \circ F(f) = F(g \circ f) = F(\text{id}_a)$ should be the identity on $F(a)$.

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  • $\begingroup$ that's pretty cool. the right diagram (imagining an F(gf) there in your picture) would commute since F(gf) = F(g)F(f) $\endgroup$ – Coreymonsta Aug 15 '18 at 20:04

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