hcju/stacksqa huggingface dataset:
We first prove this when $K/F$ is purely inseparable. Namely, we claim that in this case there is a unique map $K \to \overline{F}$. This can be seen by choosing a sequence of elements $\alpha_1, \ldots, \alpha_n \in K$ as in Lemma \ref{lemma-finite-purely-inseparable}. The irreducible polynomial of $\alpha_i$ over $F(\alpha_1, \ldots, \alpha_{i - 1})$ is $x^p - \alpha_i^p$. Applying Lemma \ref{lemma-count-embeddings-explicitly} we see that $|\Mor_F(K, \overline{F})| = 1$. On the other hand, $[K : F]_s = 1$ in this case hence the equality holds. \medskip\noindent Let's return to a general finite extension $K/F$. In this case choose $F \subset K_s \subset K$ as in Lemma \ref{lemma-separable-first}. By Lemma \ref{lemma-separable-equality} we have $|\Mor_F(K_s, \overline{F})| = [K_s : F] = [K : F]_s$. On the other hand, every field map $\sigma' : K_s \to \overline{F}$ extends to a unique field map $\sigma : K \to \overline{F}$ by the result of the previous paragraph. In other words $|\Mor_F(K, \overline{F})| = |\Mor_F(K_s, \overline{F})|$ and the proof is done.
We first prove this in case $K$ is finite over $F$. Since we have multiplicativity for the usual degree (by Lemma \ref{lemma-multiplicativity-degrees}) it suffices to prove one of the two formulas. By Lemma \ref{lemma-separable-degree} we have $[K : F]_s = |\Mor_F(K, \overline{F})|$. By the same lemma, given any $\sigma \in \Mor_F(E, \overline{F})$ the number of extensions of $\sigma$ to a map $\tau : K \to \overline{F}$ is $[K : E]_s$. Namely, via $E \cong \sigma(E) \subset \overline{F}$ we can view $\overline{F}$ as an algebraic closure of $E$. Combined with the fact that there are $[E : F]_s = |\Mor_F(E, \overline{F})|$ choices for $\sigma$ we obtain the result. \medskip\noindent If the extensions are infinite one can write $K$ as the union of all finite subextension $F \subset K' \subset K$. For each $K'$ we set $E' = E \cap K'$. Then we have the formulas of the lemma for $K'/E'/F$ by the first paragraph. Since $[K : F]_s = \sup \{[K' : F]_s\}$ and similarly for the other degrees (some details omitted) we obtain the result in general.
Let $\mathcal{P}$ be the set of all minimal polynomials over $F$ of all elements of $E$. Set $$ T = \{\beta \in \overline{F} \mid P(\beta) = 0\text{ for some }P \in \mathcal{P}\} $$ It is clear that if $E$ is normal over $F$, then $\sigma(E) = T$ for all $\sigma \in \Mor_F(E, \overline{F})$. Thus we see that (1) implies (2). \medskip\noindent Conversely, assume (2). Pick $\beta \in T$. We can find a corresponding $\alpha \in E$ whose minimal polynomial $P \in \mathcal{P}$ annihilates $\beta$. Because $F(\alpha) = F[x]/(P)$ we can find an element $\sigma_0 \in \Mor_F(F(\alpha), \overline{F})$ mapping $\alpha$ to $\beta$. By Lemma \ref{lemma-map-into-algebraic-closure} we can extend $\sigma_0$ to a $\sigma \in \Mor_F(E, \overline{F})$. Whence we see that $\beta$ is in the common image of all embeddings $\sigma : E \to \overline{F}$. It follows that $\sigma(E) = T$ for any $\sigma$. Fix a $\sigma$. Now let $P \in \mathcal{P}$. Then we can write $$ P = (x - \beta_1) \ldots (x - \beta_n) $$ for some $n$ and $\beta_i \in \overline{F}$ by Lemma \ref{lemma-algebraically-closed}. Observe that $\beta_i \in T$. Thus $\beta_i = \sigma(\alpha_i)$ for some $\alpha_i \in E$. Thus $P = (x - \alpha_1) \ldots (x - \alpha_n)$ splits completely over $E$. This finishes the proof.
Let $P_i$ be the minimal polynomial of $\alpha_i$ over $F$. Let $\alpha_i = \alpha_{i, 1}, \alpha_{i, 2}, \ldots, \alpha_{i, d_i}$ be the roots of $P_i$ over $E$. Given two embeddings $\sigma, \sigma' : E \to \overline{F}$ over $F$ we see that $$ \{\sigma(\alpha_{i, 1}), \ldots, \sigma(\alpha_{i, d_i})\} = \{\sigma'(\alpha_{i, 1}), \ldots, \sigma'(\alpha_{i, d_i})\} $$ because both sides are equal to the set of roots of $P_i$ in $\overline{F}$. The elements $\alpha_{i, j}$ generate $E$ over $F$ and we find that $\sigma(E) = \sigma'(E)$. Hence $E/F$ is normal by Lemma \ref{lemma-characterize-normal}.
Choose an algebraic closure $\overline{L}$ of $L$ (Theorem \ref{theorem-existence-algebraic-closure}). \medskip\noindent Let $\tau$ be as in (1). Then $\tau(M) = M$ as subfields of $\overline{L}$ by Lemma \ref{lemma-characterize-normal} and hence $\tau|_M : M \to M$ is an automorphism. \medskip\noindent Let $\sigma : M \to L$ be as in (2). By Lemma \ref{lemma-map-into-algebraic-closure} we can extend $\sigma$ to a map $\tau : L \to \overline{L}$, i.e., such that $$ \xymatrix{ L \ar[r]_\tau & \overline{L} \\ M \ar[u] \ar[ru]_\sigma & K \ar[l] \ar[u] } $$ is commutative. By Lemma \ref{lemma-characterize-normal} we see that $\tau(L) = L$. Hence $\tau : L \to L$ is an automorphism which extends $\sigma$.
Choose an algebraic closure $\overline{F}$ of $F$. Recall that $[E : F]_s = |\Mor_F(E, \overline{F})|$. Pick an element $\sigma_0 \in \Mor_F(E, \overline{F})$. Then the map $$ \text{Aut}(E/F) \longrightarrow \Mor_F(E, \overline{F}),\quad \tau \longmapsto \sigma_0 \circ \tau $$ is injective. Thus the inequality. If equality holds, then every $\sigma \in \Mor_F(E, \overline{F})$ is gotten by precomposing $\sigma_0$ by an automorphism. Hence $\sigma(E) = \sigma_0(E)$. Thus $E$ is normal over $F$ by Lemma \ref{lemma-characterize-normal}. \medskip\noindent Conversely, assume that $E/F$ is normal. Then by Lemma \ref{lemma-characterize-normal} we have $\sigma(E) = \sigma_0(E)$ for all $\sigma \in \Mor_F(E, \overline{F})$. Thus we get an automorphism of $E$ over $F$ by setting $\tau = \sigma_0^{-1} \circ \sigma$. Whence the map displayed above is surjective.
Choose an algebraic closure $\overline{F}$. Then we can write $P = c (x - \beta_1) \ldots (x - \beta_n)$ in $\overline{F}[x]$, see Lemma \ref{lemma-algebraically-closed}. Note that $c \in F^*$. Set $E = F(\beta_1, \ldots, \beta_n)$. Then it is clear that $E$ is minimal with the requirement that $P$ splits completely over $E$. \medskip\noindent Next, let $E'$ be another minimal field extension of $F$ such that $P$ splits completely over $E'$. Write $P = c (x - \alpha_1) \ldots (x - \alpha_n)$ with $c \in F$ and $\alpha_i \in E'$. Again it follows from minimality that $E' = F(\alpha_1, \ldots, \alpha_n)$. Moreover, if we pick any $\sigma : E' \to \overline{F}$ (Lemma \ref{lemma-map-into-algebraic-closure}) then we immediately see that $\sigma(\alpha_i) = \beta_{\tau(i)}$ for some permutation $\tau : \{1, \ldots, n\} \to \{1, \ldots, n\}$. Thus $\sigma(E') = E$. This implies that $E'$ is a normal extension of $F$ by Lemma \ref{lemma-characterize-normal} and that $E \cong E'$ as extensions of $F$ thereby finishing the proof.
Choose generators $\alpha_1, \ldots, \alpha_n$ of $E$ over $F$. Let $P_1, \ldots, P_n$ be the minimal polynomials of $\alpha_1, \ldots, \alpha_n$ over $F$. Set $P = P_1 \ldots P_n$. Observe that $(x - \alpha_1) \ldots (x - \alpha_n)$ divides $P$, since each $(x - \alpha_i)$ divides $P_i$. Say $P = (x - \alpha_1) \ldots (x - \alpha_n)Q$. Let $K/E$ be the splitting field of $P$ over $E$. We claim that $K$ is the splitting field of $P$ over $F$ as well (which implies that $K$ is normal over $F$). This is clear because $K/E$ is generated by the roots of $Q$ over $E$ and $E$ is generated by the roots of $(x - \alpha_1) \ldots (x - \alpha_n)$ over $F$, hence $K$ is generated by the roots of $P$ over $F$. \medskip\noindent Uniqueness. Suppose that $K'/E$ is a second smallest extension such that $K'/F$ is normal. Choose an algebraic closure $\overline{F}$ and an embedding $\sigma_0 : E \to \overline{F}$. By Lemma \ref{lemma-map-into-algebraic-closure} we can extend $\sigma_0$ to $\sigma : K \to \overline{F}$ and $\sigma' : K' \to \overline{F}$. By Lemma \ref{lemma-intersect-normal} we see that $\sigma(K) \cap \sigma'(K')$ is normal over $F$. By minimality we conclude that $\sigma(K) = \sigma(K')$. Thus $\sigma \circ (\sigma')^{-1} : K' \to K$ gives an isomorphism of extensions of $E$.
Proof of (1). Let $M'$ be the smallest subextension of $L/K$ containing $M$ which is normal over $K$. By Lemma \ref{lemma-normal-closure} this is the normal closure of $M/K$ and is finite over $K$. \medskip\noindent Proof of (2). Let $\alpha_1, \ldots, \alpha_n \in M'$ generate $M'$ over $M$. Let $P_1, \ldots, P_n$ be the minimal polynomials of $\alpha_1, \ldots, \alpha_n$ over $K$. Let $\alpha_{i, j}$ be the roots of $P_i$ in $L$. Let $M''  = M(\alpha_{i, j})$. It follows from Lemma \ref{lemma-normally-generated} (applied with the set of generators $M \cup \{\alpha_{i, j}\}$) that $M''$ is normal over $K$.
Choose an algebraic closure $\overline{K}$ of $K$. Set $n = [L : K]_s = |\Mor_K(L, \overline{K})|$ with equality by Lemma \ref{lemma-separable-degree}. Say $\Mor_K(L, \overline{K}) = \{\sigma_1, \ldots, \sigma_n\}$. Let $M' \subset \overline{K}$ be the $K$-subalgebra generated by $\sigma_i(L)$, $i = 1, \ldots, n$. It follows from Lemma \ref{lemma-characterize-normal} that $M'$ is normal over $K$ and that it is the smallest normal subextension of $\overline{K}$ containing $\sigma_1(L)$. By uniqueness of normal closure we have $M \cong M'$. Finally, there is a surjective map $$ L \otimes_K L \otimes_K \ldots \otimes_K L \longrightarrow M', \quad \lambda_1 \otimes \ldots \otimes \lambda_n \longmapsto \sigma_1(\lambda_1) \ldots \sigma_n(\lambda_n) $$ and note that $n \leq [L : K]$ by definition.
Let $\alpha \in E$ be a primitive element. Let $P$ be the minimal polynomial of $\alpha$ over $F$. Let $E \subset M$ be a splitting field for $P$ over $E$, so that $P(x) = (x - \alpha)(x - \alpha_2) \ldots (x - \alpha_n)$ over $M$. For ease of notation we set $\alpha_1 = \alpha$. Next, let $E/K/F$ be a subextension. Let $Q$ be the minimal polynomial of $\alpha$ over $K$. Observe that $\deg(Q) = [E : K]$. Writing $Q = x^d + \sum_{i < d} a_i x^i$ we claim that $K$ is equal to $L = F(a_0, \ldots, a_{d - 1})$. Indeed $\alpha$ has degree $d$ over $L$ and $L \subset K$. Hence $[E : L] = [E : K]$ and it follows that $[K : L] = 1$, i.e., $K = L$. Thus it suffices to show there are at most finitely many possibilities for the polynomial $Q$. This is clear because we have a factorization $P = QR$  in $K[x]$ in particular in $E[x]$. Since we have unique factorization in $E[x]$ there are at most finitely many monic factors of $P$ in $E[x]$. \medskip\noindent If $F$ is a finite field (equivalently $E$ is a finite field), then $E/F$ has a primitive element by the discussion in Section \ref{section-finite}. Next, assume $F$ is infinite and there are at most finitely many proper subfields $E/K/F$. List them, say $K_1, \ldots, K_N$. Then each $K_i \subset E$ is a proper sub $F$-vector space. As $F$ is infinite we can find a vector $\alpha \in E$ with $\alpha \not \in K_i$ for all $i$ (a vector space can never be equal to a finite union of proper subvector spaces; details omitted). Then $\alpha$ is a primitive element for $E$ over $F$. \medskip\noindent Having established the equivalence of (1) and (2) we now turn to the final statement of the lemma. Choose an algebraic closure $\overline{F}$ of $F$. Enumerate the elements $\sigma_1, \ldots, \sigma_n \in \Mor_F(E, \overline{F})$. Since $E/F$ is separable we have $n = [E : F]$ by Lemma \ref{lemma-separable-equality}. Note that if $i \not = j$, then $$ V_{ij} = \Ker(\sigma_i - \sigma_j : E \longrightarrow \overline{F}) $$ is not equal to $E$. Hence arguing as in the preceding paragraph we can find $\alpha \in E$ with $\alpha \not \in V_{ij}$ for all $i \not = j$. It follows that $|\Mor_F(F(\alpha), \overline{F})| \geq n$. On the other hand $[F(\alpha) : F] \leq [E : F]$. Hence equality by Lemma \ref{lemma-separable-equality} and we conclude that $E = F(\alpha)$.
Choose a basis $\beta_1, \ldots, \beta_e$ of $L$ over $K(\alpha)$. Then $e$ satisfies $e \deg(P) = [L : K]$ by Lemmas \ref{lemma-degree-minimal-polynomial} and \ref{lemma-multiplicativity-degrees}. Then we see that $L = \bigoplus K(\alpha) \beta_i$ is a direct sum decomposition into $\alpha$-invariant subspaces hence the characteristic polynomial of $\alpha : L \to L$ is equal to the characteristic polynomial of $\alpha : K(\alpha) \to K(\alpha)$ to the power $e$. \medskip\noindent To finish the proof we may assume that $L = K(\alpha)$. In this case by Cayley-Hamilton we see that $\alpha$ is a root of the characteristic polynomial. And since the characteristic polynomial has the same degree as the minimal polynomial, we find that equality holds.
It is clear that (3) implies (2). If (2) holds, then pick $\gamma \in L$ with $\text{Trace}_{L/K}(\gamma) \not = 0$. Then if $\alpha \in L$ is nonzero, we see that $Q_{L/K}(\alpha, \gamma/\alpha) \not = 0$. Hence $Q_{L/K}$ is nondegenerate. This proves the equivalence of (2) and (3). \medskip\noindent Suppose that $K$ has characteristic $p$ and $L = K(\alpha)$ with $\alpha \not \in K$ and $\alpha^p \in K$. Then $\text{Trace}_{L/K}(1) = p = 0$. For $i = 1, \ldots, p - 1$ we see that $x^p - \alpha^{pi}$ is the minimal polynomial for $\alpha^i$ over $K$ and we find $\text{Trace}_{L/K}(\alpha^i) = 0$ by Lemma \ref{lemma-trace-and-norm-from-minimal-polynomial}. Hence for this kind of purely inseparable degree $p$ extension we see that $\text{Trace}_{L/K}$ is identically zero. \medskip\noindent Assume that $L/K$ is not separable. Then there exists a subfield $L/K'/K$ such that $L/K'$ is a purely inseparable degree $p$ extension as in the previous paragraph, see Lemmas \ref{lemma-separable-first} and \ref{lemma-finite-purely-inseparable}. Hence by Lemma \ref{lemma-trace-and-norm-tower} we see that $\text{Trace}_{L/K}$ is identically zero. \medskip\noindent Assume on the other hand that $L/K$ is separable. By induction on the degree we will show that $\text{Trace}_{L/K}$ is not identically zero. Thus by Lemma \ref{lemma-trace-and-norm-tower} we may assume that $L/K$ is generated by a single element $\alpha$ (use that if the trace is nonzero then it is surjective). We have to show that $\text{Trace}_{L/K}(\alpha^e)$ is nonzero for some $e \geq 0$. Let $P = x^d + a_1 x^{d - 1} + \ldots + a_d$ be the minimal polynomial of $\alpha$ over $K$. Then $P$ is also the characteristic polynomial of the linear maps $\alpha : L \to L$, see Lemma \ref{lemma-characteristic-vs-minimal-polynomial}. Since $L/k$ is separable we see from Lemma \ref{lemma-recognize-separable} that $P$ has $d$ pairwise distinct roots $\alpha_1, \ldots, \alpha_d$ in an algebraic closure $\overline{K}$ of $K$. Thus these are the eigenvalues of $\alpha : L \to L$. By linear algebra, the trace of $\alpha^e$ is equal to $\alpha_1^e + \ldots + \alpha_d^e$. Thus we conclude by Lemma \ref{lemma-sums-of-powers}.
Assume $|\text{Aut}(E/F)| = [E : F]$. By Lemma \ref{lemma-normal-and-automorphisms} this implies that $E/F$ is separable and normal, hence Galois. Conversely, if $E/F$ is separable then $[E : F] = [E : F]_s$ and if $E/F$ is in addition normal, then Lemma \ref{lemma-normal-and-automorphisms} implies that $|\text{Aut}(E/F)| = [E : F]$.
Given $\alpha \in K$ consider the orbit $G \cdot \alpha \subset K$ of $\alpha$ under the group action. Consider the polynomial $$ P = \prod\nolimits_{\beta \in G \cdot \alpha} (x - \beta) \in K[x] $$ The key to the whole lemma is that this polynomial is invariant under the action of $G$ and hence has coefficients in $K^G$. Namely, for $\tau \in G$ we have $$ P^\tau = \prod\nolimits_{\beta \in G \cdot \alpha} (x - \tau(\beta)) = \prod\nolimits_{\beta \in G \cdot \alpha} (x - \beta) = P $$ because the map $\beta \mapsto \tau(\beta)$ is a permutation of the orbit $G \cdot \alpha$. Thus $P \in K^G[x]$. Since also $P(\alpha) = 0$ as $\alpha$ is an element of its orbit we conclude that the extension $K/K^G$ is algebraic. Moreover, the minimal polynomial $Q$ of $\alpha$ over $K^G$ divides the polynomial $P$ just constructed. Hence $Q$ is separable (by Lemma \ref{lemma-recognize-separable} for example) and we conclude that $K/K^G$ is separable. Thus $K/K^G$ is Galois. To finish the proof it suffices to show that $[K : K^G] = |G|$ since then $G$ will be the Galois group by Lemma \ref{lemma-finite-Galois}. \medskip\noindent Pick finitely many elements $\alpha_i \in K$, $i = 1, \ldots, n$ such that $\sigma(\alpha_i) = \alpha_i$ for $i = 1, \ldots, n$ implies $\sigma$ is the neutral element of $G$. Set $$ L = K^G(\{\sigma(\alpha_i); 1 \leq i \leq n, \sigma \in G\}) \subset K $$ and observe that the action of $G$ on $K$ induces an action of $G$ on $L$. We will show that $L$ has degree $|G|$ over $K^G$. This will finish the proof, since if $L \subset K$ is proper, then we can add an element $\alpha \in K$, $\alpha \not \in L$ to our list of elements $\alpha_1, \ldots, \alpha_n$ without increasing $L$ which is absurd. This reduces us to the case that $K/K^G$ is finite which is treated in the next paragraph. \medskip\noindent Assume $K/K^G$ is finite. By Lemma \ref{lemma-primitive-element} we can find $\alpha \in K$ such that $K = K^G(\alpha)$. By the construction in the first paragraph of this proof we see that $\alpha$ has degree at most $|G|$ over $K$. However, the degree cannot be less than $|G|$ as $G$ acts faithfully on $K^G(\alpha) = L$ by construction and the inequality of Lemma \ref{lemma-normal-and-automorphisms}.
Namely, by Lemma \ref{lemma-lift-maps} we see that every element $\tau \in \text{Gal}(L/K)$ induces an element $\tau|_M \in \text{Gal}(M/K)$ which gives us the homomorphism on the right. The map on the left identifies the left group with the kernel of the right arrow. The sequence is exact because the sizes of the groups work out correctly by multiplicativity of degrees in towers of finite extensions (Lemma \ref{lemma-multiplicativity-degrees}). One can also use Lemma \ref{lemma-lift-maps} directly to see that the map on the right is surjective.
Throughout this proof we think of $E$ as a discrete topological space. Recall that the compact open topology on the set of self maps $\text{Map}(E, E)$ is the universal topology such that the action $\text{Map}(E, E) \times E \to E$ is continuous. See Topology, Example \ref{topology-example-automorphisms-of-a-set} for a precise statement. The topology of the lemma on $\text{Gal}(E/F)$ is the induced topology coming from the injective map $\text{Gal}(E/F) \to \text{Map}(E, E)$. Hence the universal property (1) follows from the corresponding universal property of the compact open topology. Since the set of invertible self maps $\text{Aut}(E)$ endowed with the compact open topology forms a topological group, see Topology, Example \ref{topology-example-automorphisms-of-a-set}, and since $\text{Gal}(E/F) = \text{Aut}(E/F) \to \text{Map}(E, E)$ factors through $\text{Aut}(E)$ we obtain a topological group. In other words, we are using the injection $$ \text{Gal}(E/F) \subset \text{Aut}(E) $$ to endow $\text{Gal}(E/F)$ with the induced structure of a topological group (see Topology, Section \ref{topology-section-topological-groups}) and by construction this is the coarsest structure of a topological group such that the action $\text{Gal}(E/F) \times E \to E$ is continuous. \medskip\noindent To show that $\text{Gal}(E/F)$ is profinite we argue as follows (our argument is necessarily nonstandard because we have defined the topology before showing that the Galois group is an inverse limit of finite groups). By Topology, Lemma \ref{topology-lemma-profinite-group} it suffices to show that the underlying topological space of $\text{Gal}(E/F)$ is profinite. For any subset $S \subset E$ consider the set $$ G(S) = \{ f : S \to E \mid \begin{matrix} f(\alpha)\text{ is a root of the minimal polynomial}\\ \text{of }\alpha\text{ over }F\text{ for all }\alpha \in S \end{matrix} \} $$ Since a polynomial has only a finite number of roots we see that $G(S)$ is finite for all $S \subset E$ finite. If $S \subset S'$ then restriction gives a map $G(S') \to G(S)$. Also, observe that if $\alpha \in S \cap F$ and $f \in G(S)$, then $f(\alpha) = \alpha$ because the minimal polynomial is linear in this case. Consider the profinite topological space $$ G = \lim_{S \subset E\text{ finite}} G(S) $$ Consider the canonical map $$ c : \text{Gal}(E/F) \longrightarrow G,\quad \sigma \longmapsto (\sigma|_S : S \to E)_S $$ This is injective and unwinding the definitions the reader sees the topology on $\text{Gal}(E/F)$ as defined above is the induced topology from $G$. An element $(f_S) \in G$ is in the image of $c$ exactly if (A) $f_S(\alpha) + f_S(\beta) = f_S(\alpha + \beta)$ and (M) $f_S(\alpha)f_S(\beta) = f_S(\alpha\beta)$ whenever this makes sense (i.e., $\alpha, \beta, \alpha + \beta, \alpha\beta \in S$). Namely, this means $\lim f_S : E \to E$ will be an $F$-algebra map and hence an automorphism by Lemma \ref{lemma-algebraic-extension-self-map}. The conditions (A) and (M) for a given triple $(S, \alpha, \beta)$ define a closed subset of $G$ and hence $\text{Gal}(E/F)$ is homeomorphic to a closed subset of a profinite space and therefore profinite itself.
By Lemma \ref{lemma-lift-maps} given $\tau : L \to L$ in $\text{Gal}(L/K)$ the restriction $\tau|_M : M \to M$ is an element of $\text{Gal}(M/K)$. This defines the homomorphism $c$. Continuity follows from the universal property of the topology: the action $$ \text{Gal}(L/K) \times M \longrightarrow M,\quad (\tau, x) \longmapsto \tau(x) = c(\tau)(x) $$ is continuous as $M \subset L$ and the action $\text{Gal}(L/K) \times L \to L$ is continuous. Hence continuity of $c$ by part (1) of Lemma \ref{lemma-galois-profinite}. Lemma \ref{lemma-lift-maps} also shows that the map is surjective.
Every subfield of $L$ containing $K$ is separable over $K$ (follows immediately from the definition). Let $S \subset L$ be a finite subset. Then $K(S)/K$ is finite and there exists a tower $L/E/K(S)/K$ such that $E/K$ is finite Galois, see Lemma \ref{lemma-normal-closure-inside-normal}. Hence $E = L_\lambda$ for some $\lambda \in \Lambda$. This certainly implies the set $\Lambda$ is not empty. Also, given $\lambda_1, \lambda_2 \in \Lambda$ we can write $L_{\lambda_i} = K(S_i)$ for finite sets $S_1, S_2 \subset L$ (Lemma \ref{lemma-finite-finitely-generated}). Then there exists a $\lambda \in \Lambda$ such that $K(S_1 \cup S_2) \subset L_\lambda$. Hence $\lambda \geq \lambda_1, \lambda_2$ and $\Lambda$ is directed (Categories, Definition \ref{categories-definition-directed-system}). Finally, since every element in $L$ is contained in $L_\lambda$ for some $\lambda \in \Lambda$, it follows from the description of filtered colimits in Categories, Section \ref{categories-section-directed-colimits} that $\colim L_\lambda = L$. \medskip\noindent If $\lambda \geq \lambda'$ in $\Lambda$, then we obtain a canonical surjective map $G_\lambda \to G_{\lambda'}$, $\sigma \mapsto \sigma|_{L_{\lambda'}}$ by Lemma \ref{lemma-ses-galois}. Thus we get an inverse system of finite groups with surjective transition maps. \medskip\noindent Recall that $G = \text{Aut}(L/K)$. By Lemma \ref{lemma-galois-infinite} the restriction $\sigma|_{L_\lambda}$ of a $\sigma \in G$ to $L_\lambda$ is an element of $G_\lambda$. Moreover, this procedure gives a continuous surjection $G \to G_\lambda$. Since the transition mappings in the inverse system of $G_\lambda$ are given by restriction also, it is clear that we obtain a canonical continuous map $$ G \longrightarrow \lim_{\lambda \in \Lambda} G_\lambda $$ Continuity by definition of limits in the category of topological groups; recall that these limits commute with the forgetful functor to the categories of sets and topological spaces by Topology, Lemma \ref{topology-lemma-topological-group-limits}. On the other hand, since $L = \colim L_\lambda$ it is clear that any element of the inverse limit (viewed as a set) defines an automorphism of $L$. Thus the map is bijective. Since the topology on both sides is profinite, and since a bijective continuous map of profinite spaces is a homeomorphism (Topology, Lemma \ref{topology-lemma-bijective-map}), the proof is complete.