How are the words "extend", "extending", "extension" used in these particular math contexts. In the following three examples, it has the words: extends, extension or extending.  I often see these three words in the context of mapping. Can someone tell me how these words are used and what they mean in the three particular context.  Is it related to the concept of function extension or group or field extension, etc.  I put all three examples in one post since the three words are close in meaning. I place the word in boldface in all three cases.
Example 1:
A group $F$ is said to be free on a subset $X\subset F$ if, given any group $G$ and any map $f:X\rightarrow G$, there is a unique homomorphism $f':F\rightarrow G$ $\textbf{extending}$ $f$, that is, having the property that $f'(x)=f(x)$ for all $x\in X.$  or the following maps
$\text{inc}:X\rightarrow F,$ $f:X\rightarrow G,$ and $\exists !f':F \dashrightarrow G$
are commutative.  Then $X$ is called a basis of $F$ and $|X|$ the rank of $F$, written $r(F).$
Example 2:
Deduce that
$S_{6}=\{{t'}_{1},\ldots,{t'}_{5}\}$ and that the map
$(1 2)\mapsto {t'}_{1}$
$(2 3)\mapsto {t'}_{2}$
$(3 4)\mapsto {t'}_{3}$
$(4 5)\mapsto {t'}_{4}$
$(5 6)\mapsto {t'}_{5}$
$\textbf{extends}$ to an automorphism of $S_6$
$\text{Example 3:}$
Let $G$ be an $\textbf{extension}$ of $A_6$ by an exterior automorphism of $S_6$ of order $2$. How does this group look like, is it just $S_6$, or where is the difference?
Thank you in advance
 A: Examples 1 and 2 are similiar, in the sense that you have two sets $A$ and $B$, a function $f\colon A_0 \to B$ from a subset $A_0$ of $A$ to $B$ and you're asking when you can find function $g\colon A \to B$ defined on the whole set $A$ such that $g(a)$ and $f(a)$ coincide for every $a \in A_0$. In this scenario, we say that the function $g$ is an extension of the function $f$ to the whole set $A$, or that $f$ was extended to $A$. Also, in both examples, one asks more than an simple extension of functions: $A$ and $B$ both have the structure of a group and while $f$ was a mere function the extension $g$ is a homomorphism.
Example 3 talks about extensions of groups. The nomenclature for such extensions is not uniform and different authors may exchange the roles of the groups $N$ and $Q$ in what follows, but let's say for now that a group $G$ is an extension of $N$ by $Q$ if $G$ has a normal subgroup $M$ isomorphic $N$ such that the quotient group $G/M$ is isomorphic to $Q$. At first, this talks about a different kind of "extension" that is unrelated to extending functions from subsets of it's desired domain as in Examples 1 and 2. The idea is that $G$ is a group that can be put together by combining the groups $N$ and $Q$ -- such as taking direct products $N \times Q$, semi-direct products $N \rtimes Q$, but there may be many other types extensions.
Now this is guessing, but I guess one reason this is called a group extension is because such $G$ is in bijection (as sets!) with the set $M \times G/M$, which in turn is in bijection with the set $N \times Q$. To see this, choose a complete set of representatives $T$ of the cosets $G/M$. This set is in bijection with $Q$. Since every element of $G$ can be uniquely written as $g = mt$ for $m \in M$ and $t \in T$, the bijection follows. Now, we already have a group structure on the subsets $N \times \{1\}$ and $\{1\} \times Q$ of $N \times Q$, and a group extension amounts to extending this group structure to the whole set $N \times Q$, that is, one wants to extend the multiplication map $$-\cdot -\colon (N \times \{1\})^2 \cup (\{1\} \times Q)^2 \to N \times Q$$ to the whole set $(N\times Q)^2$ in such a way that it satisfy the group axioms. This may be a stretch, but it's one way that the meanings coincide. I can think of other historical reasons why it may be called an extension as well.
Now for your last question, I guess your construction means this: take any non-inner automorphism $f \in \text{Aut}(S_6)$ and consider the subgroup of $\text{Aut}(S_6)$ generated by $f$. Since $f$ has order two, that is, $f\circ f=\text{id}$, this group is isomorphic to the cyclic group $\mathbb{Z}/2\mathbb{Z}$ of order $2$. Moreover, since $A_6$ is a characteristic subgroup of $S_6$ (it's equal to it's commutator subgroup), we have $f(A_6) \subseteq A_6$, that is, $f$ also induces an automorphism of $A_6$. This allows us to construct the semidirect product $A_6 \rtimes_f \mathbb{Z}/2\mathbb{Z}$, which is the set $A_6 \times \mathbb{Z}/2\mathbb{Z}$ with the group structure given by $$(\sigma_1,m)\cdot(\sigma_2,n) = (\sigma_1 f^m(\sigma_2), m + n)\,.$$ Now the question of whether this group is isomorphic to $S_6$ is an interesting one. The fact that you are considering $S_6$ and not $S_n$ for any other choice of $n$ is very peculiar, but it is interesting to note that $S_n$ is indeed always of the form $A_n \rtimes_f \mathbb{Z}/2\mathbb{Z}$ for $n \geq 5$ and some automorphism $f$ of $A_n$ -- try to figure out what automorphism this is, starting by look at the automorphism groups of symmetric and alternating groups.
