**1. About injectivity:** Yves already answered about injectivity in the comments. Below is an alternative argument which works more generally for free products:

**Proposition 1:** *Let $G=A_1 \ast \cdots \ast A_n$ be a free products and let $w_1, \ldots, w_n \in G$ be $n$ elements. Then $$H:=\langle w_1A_1w_1, \ldots, w_nA_nw_n^{-1} \rangle = w_1A_1w_1^{-1} \ast \cdots \ast w_nA_nw_n^{-1}.$$*

**Sketch of proof.** Think of $G$ as the fundamental group of the graph of groups having a central vertex labelled by $\{1\}$ linked to $n$ vertices labelled by $A_1, \ldots, A_n$ and whose edges are all labelled by $\{1\}$. Let $T$ denote the corresponding Bass-Serre tree.

Choose $z_1, \ldots, z_n \in G$ such that $H= \langle z_1A_1z_1^{-1}, \ldots, z_nA_nz_n^{-1} \rangle$ with $|z_1| + \cdots + |z_n|$ minimal. For every $1 \leq i \leq n$, let $v_i$ denote the vertex of $T$ stabilised by $z_i A_i z_i^{-1}$. Also, let $v$ denote the vertex of $T$ corresponding to the coset $\{1\}$.

I claim that, for every $i \neq j$, $v_i$ does not belong to $[v_j,v]$. Otherwise, take a $z \in \mathrm{stab}(v_i)$ sending the first edge of $[v_i,v_j]$ to the first edge of $[v_i,v]$. Then the length of $zz_j$ is smaller than the length of $z_j$ because $z \cdot v_j$ is closer to $v$ than $v_j$. This contradicts the choice of the $z_i$'s.

As a consequence, there exists a subtree $S \subset T$ whose leaves are the $v_i$'s. Now a ping-pong argument leads to the desired conclusion. $\square$

**2. About surjectivity:** Given an $n$-tuple $(a_1, \ldots, a_n)$ of words written over the generators (and their inverses) of $\mathbb{F}_n$, consider the following elementary transformations:

- if $a_i$ is not reduced, then reduce it;
- if $x_i^{\pm 1}$ is the last letter of $a_i$, remove it;
- if $a_jx_j^{\pm 1}$ is a prefix of $a_i$, remove from $a_i$ the last letter of this prefix.

Notice that the total length $|a_1|+ \cdots +|a_n|$ decreases when an elementary transformation is applied. Call $(a_1, \ldots, a_n)$ a *reduction* of $(w_1, \ldots, w_n)$ if $(a_1, \ldots, a_n)$ is obtained from $(w_1, \ldots, w_n)$ by applying elementary transformations until it is no longer possible.

**Proposition 2:** *Let $(a_1, \ldots, a_n)$ a reduction of $(w_1, \ldots, w_n)$. Then $\varphi$ is surjective if and only if $a_1= \cdots a_n=1$.*

**Sketch of proof.** Let $\Gamma_0$ be a bouquet of balloons, such that each wire is an oriented path of length $|w_i|$ labelled by $w_i$ and its corresponding balloon is a single oriented edge labelled by $x_i$. Let $\Gamma_1, \ldots, \Gamma_k$ be a sequence of graphs such that $\Gamma_i$ is obtained from $\Gamma_{i-1}$ by applying a Stallings' fold, and such that no such a fold applies to $\Gamma_k$. Notice that $\Gamma_k$ looks like a bouquet of balloons in which the wires are glued together like a rooted tree. Let $a_1, \ldots, a_n$ denote the words labelling the oriented paths from the root to the balloons.

By looking at how a Stallings' fold works, we find that $(a_1, \ldots, a_n)$ is a reduction of $(w_1, \ldots, w_n)$.

If $\Gamma$ is a bouquet of $n$ circles, because no Stallings' fold applies to $\Gamma_k$, the canonical map $\Gamma_k \to \Gamma$ is a local isometry*. As shown by Stallings, it is possible to add edges to $\Gamma_k$ in order to get a new graph $\Psi$ such that the local isometry $\Gamma_k \to \Gamma$ extends to a covering map $\Psi \to \Gamma$. Consequently, $\varphi$ is surjective if and only if $\Psi = \Gamma_k$, which amounts to saying that the tree-like part of $\Gamma_k$ is empty; in other words $a_1= \cdots = a_n=1$. $\square$

Below is an illustration of the method for $\mathbb{F}_2= \langle a,b \rangle$, $w_1 = ab$ and $w_2=a$:

*As a consequence, $\Gamma_k \to \Gamma$ is $\pi_1$-injective, whence a third argument showing the injectivity of $\varphi$.