The extended transition function exists and is unique

Fix $q\in Q$ and a word $w=a_0{a}_1\cdots a_{n-1}$. Define a sequence of states by $s_0=q$ and $s_{i+1}=\delta (s_i,a_i)$ for $0\le i<n$, and then define ${\delta }^{\ast }(q,w):=s_n$. This gives a function $Q\times {\mathrm{\Sigma }}^{\ast }\to Q$ because the DFA transition function supplies exactly one next state at each letter.

If $w=\epsilon$, then $n=0$ and step 1.1 gives ${\delta }^{\ast }(q,\epsilon )=s_0=q$. If $w=ua$ with $u=a_0\cdots a_{n-2}$ and $a=a_{n-1}$, then the last state in the sequence is $s_n=\delta (s_{n-1},a)=\delta ({\delta }^{\ast }(q,u),a)$. So the function from step 1.1 satisfies the two clauses of [L1].

Let $\eta :Q\times {\mathrm{\Sigma }}^{\ast }\to Q$ be any other function satisfying [L1]. We prove by induction on the length of $w$ that $\eta (q,w)={\delta }^{\ast }(q,w)$ for every $q\in Q$. For $w=\epsilon$, both values are $q$ by [L1] and step 2.1.

Assume $\eta (q,u)={\delta }^{\ast }(q,u)$ for all words $u$ of length $n$, and let $w=ua$ have length $n+1$. Then [L1] gives $\eta (q,w)=\delta (\eta (q,u),a)$, and the induction hypothesis turns this into $\delta ({\delta }^{\ast }(q,u),a)={\delta }^{\ast }(q,w)$ by step 2.1.

Steps 3.1 and 3.2 show that every function satisfying [L1] agrees with the function built in step 1.1, so the extended transition function exists and is unique.