I'm reading Categories for the Working Mathematician by Saunders Mac Lane. At the section 5 from chapter 1, for a fixed category, he claims that every arrow with right inverse, is epic (right cancellable). He claims also that the converse is true in the category of Sets, but fails in the category of Groups. I tried by myself to find a pair of groups and an arrow having these properties, but I cannot find them.

I also read that a given group, regarded as a category with one element, one arrow per element of the group, and the composition of arrows representing the group product, is not a concrete category. How can I prove that?

Thanks in advance. Every help would be very appreciated.


If $f:Q_8 \to \{\pm 1\}$ is defined by $\{1,-1,i,-i\} \mapsto 1$ and $\{j,-j,k,-k\} \mapsto -1$ then $f$ is epic, but it has no right inverse, that is, there is no homomorphism $h: \{\pm 1 \} \to Q_8$ so that $f(h(-1))=-1$. This is simply because there are only two homomorphisms from $\{\pm 1\}$ to $Q_8$, $-1 \mapsto \pm 1$, but $f(\pm1)=1 \neq -1$.

An abelian example is $f:\mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}:x+4\mathbb{Z} \mapsto x+2\mathbb{Z}$. It is epic (in any concrete category containing it), but has no right inverse since there are (at most) two $h:\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$, $h(n+2\mathbb{Z})=0+4\mathbb{ZZ}$ and $h(n+2\mathbb{Z}) = 2n + 4\mathbb{Z}$. However $f(h(1+2\mathbb{Z})) = 2n + 2\mathbb{Z} = 0 +2\mathbb{Z} \neq 1+2\mathbb{Z}$ in both cases, so $f$ has no right inverse.

Most algebraic categories are like this: not being a zero-divisor is different from being a unit. Not all epics split.

I believe a one object category in which all arrows are invertible (a “group”) is always a concrete category insofar as there is a faithful functor to the category of sets. My view of the object is as the set containing the group elements, and the arrows as either the left or right multiplication maps, which makes it a concrete category as well.

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  • $\begingroup$ In *all algebraic categories, surjective homomorphisms are epi. $\endgroup$ – Martin Brandenburg Jun 11 '13 at 9:32

2) A group as an one-element category is not a concrete category (since an one-element object has only one morphism), but it is isomorphic to a concrete one (one object with $|G|$ elements).

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  • $\begingroup$ I don't think $0 \to \mathbb{Z}$ is epic. $0 \to \mathbb{Z} \xrightarrow{x\mapsto 2x} \mathbb{Z} = 0 \to \mathbb{Z} \xrightarrow{x\mapsto 3x} \mathbb{Z}$ but $\mathbb{Z} \xrightarrow{x\mapsto 2x} \mathbb{Z} \neq \mathbb{Z} \xrightarrow{x\mapsto 3x} \mathbb{Z}$ $\endgroup$ – Jack Schmidt Jun 11 '13 at 6:41
  • $\begingroup$ @Jack Schmidt: Sorry, thank you. I missed $_2$. $\endgroup$ – Boris Novikov Jun 11 '13 at 7:01
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    $\begingroup$ $0 \to \mathbb{Z}_2$ is not epic (it fails to distinguish the zero morphism and the identity morphism $\mathbb{Z}_2 \to \mathbb{Z}_2$). It is a general (but quite non-trivial) fact that epics in the category of groups are surjective, so you really need to exhibit a surjective homomorphism without right inverse. $\endgroup$ – Martin Jun 11 '13 at 8:39
  • $\begingroup$ @Martin Yes, you are right. Thank you. $\endgroup$ – Boris Novikov Jun 11 '13 at 8:56

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