Do the maps need to be surjective in the definition of a profinite group?

Question

Let $G_i, f_{ij}$ be an inverse system of topological groups where each $G_i$ is finite in the discrete topology. A profinite group is defined to be an inverse limit of such an inverse system. However, my professor seemed to assume that the $f_{ij}$ needed to be surjective (so a projective system). Is it necessary to have this assumption? Are there any advantages to assuming surjectivity?

In any case one has that if $G = \lim\limits_{\leftarrow}G_i$, then $G$ is isomorphic to the inverse limit of all quotients $G/N$, where $N$ runs through the open normal subgroups of $G$. Therefore any profinite group will be the inverse limit of a system where the maps are surjective, so I guess we can assume this property holds without loss of generality.

Answer 1

It is not necessary for the maps to be surjective in the definition. However, any profinite group is isomorphic to a limit of a projective system in which the maps are surjective. Indeed, suppose $(G_i, f_{ij})$ is a projective system. Define $G_i' = \cap_j f_{ij} (G_j)$, where the intersection is taken over all $j \to i$. Remark that if $G_j \to G_{k} \to G_i$, we have $f_{ij}(G_j) = f_{ik}(f_{kj}(G_j)) \subseteq f_{ik}(G_{k})$, so the $f_{ij}(G_j)$ get smaller as $j$ moves up the system. I'll let you prove that $(G_i', f'_{ij})$ forms a projective system, where $f'_{ij}$ are the restrictions of the $f_{ij}$'s, and that the $f'_{ij}$'s are surjective. I'll let you prove also that the inclusion $(G'_i, f_{ij}') \to (G_i, f_{ij})$ induces an isomorphism on the limits.

Answer 2

A detailed explanation of Bruno Joyal's answer:

I. For $k \geq i$, the image of the restriction $f_{ik}' = f_{ik} \mid G_k'$ is contained in $G_i'$.

Let $$g \in G_k' = \bigcap\limits_{j \geq k} f_{kj}(G_j)$$ In order to show $f_{ik}(g) \in G_i'$, we must show that for any $j \geq i$, there exists an $h \in G_j$ such that $f_{ij}(h) = f_{ik}(g)$. Since we are in a direct system, we may choose an index $s$ which is $\geq j$ and $k$. Since $g \in G_k'$, there exists a $y \in G_s$ with $f_{ks}(y) = g$. Setting $h = f_{js}(y)$, we have $$f_{ij}(h) = f_{ij}(f_{js}(y)) = f_{is}(y) = f_{ik}(f_{ks}(y)) = f_{ik}(g)$$

II. The mapping $f_{ik}':G_k' \rightarrow G_i'$ is surjective.

Let $x \in G_i'$. Then for any $j \geq i$, there exists an $x_j \in G_j$ with $f_{ij}(x_j) = x$. In particular, $f_{ik}(x_k) = x$. However, we do not know that $x_k$ actually lies in $G_k'$. If it does not, then nevertheless it suffices to find a $g \in f_{ik}^{-1}\{x\}$ which does lie in $G_k'$.

Suppose there is no such $g$. Then for every $g \in f_{ik}^{-1}\{x\}$, there exists an index $k_g \geq k$ for which $f_{kk_g}^{-1}\{g\} = \emptyset$. Since $f_{ik}^{-1}\{x\}$ is finite (we are dealing with finite groups), we can find an index $j$ which is $\geq$ all the indices $k_g$. We then have $$x = f_{ij}(x_j) = f_{ik}(f_{kj}(x_j))$$ So $f_{kj}(x_j) \in f_{ik}^{-1}\{x\}$, hence $f_{kj}(x_j)$ is equal to some $g$. But then $$g = f_{kj}(x_j) = f_{kk_g}(f_{k_gj}(x_j))$$ so $f_{k_gj}(x_j) \in f_{kk_g}^{-1}\{g\} = \emptyset$, a contradiction.

III. $(G_i', f_{ij}')$ forms an inverse system, with $\lim\limits_{\leftarrow} G_i' \cong \lim\limits_{\leftarrow} G_i$.

It's clear that $(G_i', f_{ij}')$ forms an inverse system. For its inverse limit, we use the canonical construction, namely the subset of the product group (in the product topology) $$ \mathcal G' \subseteq \prod\limits_i G_i'$$ consisting of all $(x_i)$ such that $x_i = f_{ij}'(x_j)$ whenever $i \leq j$. On the other hand, $\lim\limits_{\leftarrow} G_i$ can be taken as the product $$\mathcal G \subseteq \prod\limits_i G_i$$ again consisting of all sequences $(x_i)$ where $x_i = f_{ij}(x_j)$ for all $i \leq j$. For any such $(x_i)$, it follows that the $x_i$ actually lie in $G_i'$, hence $$\lim\limits_{\leftarrow} G_i =\mathcal G = \mathcal G' = \lim\limits_{\leftarrow} G_i'$$