Must a surjection $F_2\to F_2$ also be an injection? In other words, are there words $w_1$ and $w_2$ in the free group $F_2$ such that $F_2=\langle w_1, w_2\rangle$ but which are not a free basis for $F_2$?
I'm sure I'm missing a simple argument or a simple example.
 A: In fact, it is true even for (finitely generated) residually free groups as a consequence of the following lemma:

Lemma: Let $G_1 \overset{\varphi_1}{\twoheadrightarrow} G_2 \overset{\varphi_2}{\twoheadrightarrow} G_3 \overset{\varphi_2}{\twoheadrightarrow} \dots$ be a sequence of epimorphisms between finitely generated residually free groups. Then all but finitely many epimorphisms are isomorphisms.

Indeed, if $\varphi : G \to G$ is an epimorphism but not an isomorphism (where $G$ is finitely generated and residually free), then the sequence $G \overset{\varphi}{\twoheadrightarrow} G \overset{\varphi}{\twoheadrightarrow} G \overset{\varphi}{\twoheadrightarrow} \dots$ contradicts the lemma.
Proof of the lemma: Let $S_1=(s_1^{(1)},...,s_n^{(1)})$ be a generator set of $G_1$; then $S_k:=\varphi_k(S_1)=(s_1^{(k)},...,s_n^{(k)})$ is a generator set of $G_k$ by surjectivity of $\varphi_k$. Let $V_k$ the set of $(M_1,...,M_n) \in SL_2(\mathbb{C})^n$ such that $r(M_1,...,M_n)=\operatorname{Id}$ for every relation $r$ of $G_k$. 
Notice that $V_k$ is an affine algebraic variety in $\mathbb{C}^{4n}$ and $V_1 \supset V_2 \supset \dots$, so the sequence $(V_k)$ is eventually constant. To conclude, it is sufficient to show that if $\varphi_k : G_k \to G_{k+1}$ is not an isomorphism, then $V_{k+1} \subsetneq V_k$.
Let $r$ be a word such that $r$ over $S_{k+1}$ is a relation of $G_{k+1}$ but $r$ over $S_k$ is not a relation of $G_k$. $G_k$ being residually free, there exists a morphism $\rho : G_{k} \to \mathbb{F}_2$ such that $\rho(r) \neq 1$. Because $\mathbb{F}_2$ is a subgroup of $SL_2(\mathbb{C})$, you can suppose that $\rho : G_{k} \to SL_2(\mathbb{C})$.
Let $N_i= \rho(s_i^{(k)}) \in SL_2(\mathbb{C})$. Then $r(N_1,...,N_n)=\rho(r) \neq 1$ so $(N_1,...,N_n) \notin V_{k+1}$. If $w$ is a word over $S_k$ such that $w$ is a relation of $G_k$, then $w(N_1,...,N_n)=\rho(w)=\rho(1)=1$ so $(N_1,...,N_n) \in V_k$. $\square$
A: Here, for reference, is the proof of Mal'cev's result that every fintely generated residually finite group $G$ is Hopfian, or in other words that the only surjections $G\to G$ are the bijections.
Let $\phi:G\to G$ be a surjection, and suppose that $\phi(g)=e$ for some $g\neq e$. By surjectivity of $\phi$ we can find succesive preimages $(g_k)_{k\geq 1}$ such that $g_1 = g$ and $\phi(g_{k+1})=g_k$ for all $k\geq 1$. Let $\alpha:G\to K$ be a finite quotient such that $\alpha(g)$ is nontrivial. I claim the maps $\alpha\circ\phi^m:G\to K$ with $m\geq0$ are all distinct: indeed $\alpha\circ\phi^m$ and $\alpha\circ\phi^n$ ($n>m$) take different values on $g_{m+1}$. But if $G$ is finitely generated then it has only finitely many distinct homomorphisms $G\to K$, a contradiction.
