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Let $G$ be a group, regarded as category with one object $*$ in which each arrow is invertible. Then the category of $G$-Sets is just the category of functors from $G$ to $\mathbf{Set}$. Now I've read that an arrow $\varphi: \tau \to \tau'$ (where $\tau$ and $\tau'$ are functors from $G$ to $\mathbf{Set}$) in the category of $G$-Sets is monic if and only if $\varphi_*$ is monic in $\mathbf{Set}$. I'm having trouble seeing why the only if part of this statement is true.

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I'd like to know the answer. The naive attempt would be, if $\phi_*(x)=\phi_*(y)$, to consider the singleton set (with its unique group action) and two maps out of it mapping to $x,y$. But those maps need not respect the group action; in fact they do precisely if $x$ and $y$ are fixed points. –  wildildildlife Oct 17 '11 at 23:27

3 Answers 3

up vote 4 down vote accepted

As wildildildlife pointed out, the naive approach is to consider morphisms $\alpha_0, \alpha_1 : \sigma\to\tau$, where $\sigma$ is the “trivial” object. The mistake is that a singleton set is a “trivial” set, we instead need a “trivial” action of $G$ (= $G$-set). It is the canonical action of $G$ on itself given by the curried 2-ary operation of $G$ (like in Cayley's theorem)!

Lets denote by $|\rho|$ the carrier and by $\triangleleft_\rho$ the operation of any action $\rho$ of $G$. Lets $\sigma$ be the canonical action of $G$, then $|\sigma|=|G|$, $\forall g_0 g_1 (g_1\triangleleft_\sigma g_0 = g_1 +_G g_0)$.

For every $x\in|\tau|$ (you defined $\tau$ in your question) define a function $\psi(g):=g\triangleleft_\tau x$. By definition of the action, $g_1\triangleleft_\tau (g_0\triangleleft_\tau x) = (g_1 + g_0)\triangleleft_\tau x$, $g_1\triangleleft_\tau (\psi(g_0)) = \psi(g_1 + g_0) = \psi(g_1\triangleleft_\sigma g_0)$, then $\psi:\sigma\to\tau$ is a homomorphism of actions. $\psi(0)=0\triangleleft_\sigma x=0 + x=x$, i.e. we can have a homomorphism which maps $0$ to any $x$ we want. Proceed as in the naive approach.

$\sigma$ is a separator in the category of actions of $G$.

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Isn't that called a generator? –  Mariano Suárez-Alvarez Oct 19 '11 at 16:26
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@Mariano Suárez-Alvarez: generator=separator. –  beroal Oct 19 '11 at 16:28

The following answer prove a more general result that show an application of yoneda lemma, so I hope you'll like it.

Let $\mathcal F, \mathcal G \colon A \to \textbf{Set}$ be two functors, then given a natural transformation $\tau \colon \mathcal F \to \mathcal G$ we have that $\tau$ is monic if and only if $\tau_a$ is monic for each $a \in A$.

Let's prove this. If for each $a \in A$ we have $\tau_a$ monic clearly given a functor $\mathcal E \colon A \to \textbf{Set}$ and two natural transformation $\sigma^1, \sigma^2 \colon \mathcal E \to \mathcal F$ such that $\tau \circ \sigma^1 = \tau \circ \sigma^2$ then we have that for each $a \in A$ hold the equality $\tau_a \circ {\sigma^1}_a = \tau_a \circ {\sigma^2}_a$ and because $\tau_a$ is monic thus it follows that $\sigma^1_a= \sigma^2_a$. Because this equation holds for each $a \in A$ we have $\sigma^1=\sigma^2$.

On the other end we also have that given a $\tau \colon \mathcal F \to \mathcal G$ which is monic then by yoneda lemma for each $a \in A$ there is a natural isomorphism $\varphi \colon \text{Nat}(A(a,-),\bullet) \stackrel{\sim}{\longrightarrow} \bullet (a)$: the functors involved are the $\hom$-functor $\text{Nat}(A(a,-),\bullet) \colon \textbf{Cat}(A,\textbf{Set}) \to \textbf{Set}$, the argument being identified by the $\bullet$, and the evaluation functor $\bullet (a) \colon \textbf{Cat}(A,\textbf{Set}) \to \textbf{Set}$, which sends every functor in its value on $a$.

Via naturality of $\varphi$ we get that the equations

$$\tau_a \circ \varphi_\mathcal{F} = \varphi_\mathcal{G} \circ \text{Nat}(A(a,-),\mathcal G)$$

must hold for each $a \in A$. But because $\varphi$ is an isomorphisms this imply that $\varphi_\mathcal{H}$ is an isomorphisms in $\textbf{Set}$ for each functor $\mathcal H \colon A \to \textbf{Set}$ and so $\tau_a = \varphi_\mathcal{G} \circ \text{Nat}(A(a,-),\tau) \circ \varphi_\mathcal{F}^{-1}$. For properties of $\hom$-functors and because of the hypothesis $\tau$ is monic also $\text{Nat}(A(a,-),\tau)$ is monic and so $\tau_a$ is monic being composition of monic.

Edit: ops I've just noted that I've forgotten to solve your question: it's a corollary to this theorem in the case when $A=G$ is group seen as a category.

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Thanks to beroal for his answer. May I add the following perspective:

To prove that in a concrete category monic implies injective, one can sometimes use an identification between 'elements' and 'arrows', because 'monic' basically means injective on arrows.

  • In $\sf{Set}$, $\sf{Top}$: $Hom(\{\star\},X)\cong X$ via $f\mapsto f(\star)$.
  • In $\sf{Gr}$ and $\sf{Ab}$: $Hom(\mathbb{Z},G)\cong G$ via $f\mapsto f(1)$.
  • In $\sf{Ring}$: $Hom(\mathbb{Z}[x],R)\cong R$ via $f\mapsto f(x)$.

In other words, the forgetful functor is represented by $\{\star\},\mathbb{Z},\mathbb{Z}[x]$ (the free object on the singleton set) respectively.

The same works in $G$-$\sf{Set}$ by taking $G$ as 'Cayley' $G$-set:

$Hom(G,X)\cong X$ via $f\mapsto f(e)$.

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And naturality of the isomorphism between a forgetful functor and a covariant hom functor implies that monic implies injective. Due to your answer I see now that my answer is not full. BTW, does a separator fit in this framework? –  beroal Oct 26 '11 at 0:23

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