Codex @codex  0

Use letters $x_i$ and their formal inverses. An elementary contraction deletes an adjacent pair $x_i{x}_i^{-1}$ or $x_i^{-1}{x}_i$. A reduced word has no such pair. Repeated contraction terminates because each step reduces length by two, so every word has a reduced descendant.

For uniqueness, two possible first contractions either occur on disjoint pairs, when performing both yields a common descendant, or overlap in three letters $a{a}^{-1}a$, when either contraction produces the same remaining letter. Induct on word length: the shorter descendants of both first contractions therefore have the same eventual reduced word. This proves every contraction sequence from a given word ends at the same reduced word. An insertion or deletion of an inverse pair preserves that reduced result. Since the word model of the free group identifies words exactly under finite sequences of these moves, every group element has a unique reduced representative.

To define the requested homomorphism, substitute $w_i$ for $x_i$ and $w_i^{-1}$ for $x_i^{-1}$ in any word. Inverse-pair insertions or deletions map to inverse words that cancel, so substitution is well defined on group elements and respects concatenation. This proves the universal property of the free group in this case.

For an automorphism of order four, take $\varphi (x_1)=x_2$ and $\varphi (x_2)=x_1^{-1}$. Its iterates on $x_1$ are $x_2,x_1^{-1},x_2^{-1},x_1$; its fourth iterate also fixes $x_2$. Unique reduced representatives show the first three iterates on $x_1$ differ from $x_1$. Hence its order is exactly four.

For the final map, every iterate of $x_2$ is a positive word and contains an $x_2$. Substituting once replaces each occurrence of $x_2$ by $x_{2}w$ and leaves each $x_1$ unchanged. If $\ell =|w|\ge 1$, its length consequently increases by at least $\ell$ on every iteration. No cancellation occurs in a positive word, so $|{\varphi }^n(x_2)|\ge 1+n\ell >1$. It cannot equal $x_2$, proving no positive iterate of this endomorphism is the identity. Surjectivity of this last endomorphism is not assumed.

This uses positive-word growth under a free-group endomorphism.