Let me give a more precise form of the exercise:
Construct infinitely many nonhomotopic retractions $S^1\vee S^1\to S^1$.
(Actually this is exercise 1.1.17 in Hatcher's Algebraic Topology)
The family of retractions, say $r_n$, which are identity on the first circle and map the second circle to go around the first circle $n$ times would suffice.
If I asked why they are nonhomotopic, most of you may argue like this:
Assume not. Then it implies that there are two retractions, say $r_n$ and $r_m$, which are homotopic restricted to the second circle. Then they must induce the same homomorphism on the level of fundamental groups. But $r_n$ maps the generator of the fundamental group of the circle to $n$ times the generator, which is a contradiction.
But a prudent reader may notice that a retraction (of $X$ to $A$, say) is a map $X\to X$, not a map with the target space $A$. (Even though the image restricts to the subspace $A$...)
Thus two retractions may be homotopic through maps from $X$ to $X$ but not through maps from $X$ to $A$. Of course it doesn't affect the argument above: the two retractions $r_n$ and $r_m$ induces the homomorphisms $\pi_1 S^1\to \pi_1(S^1\vee S^1)$, but they are mapped into the subgroup $\pi_1 S^1<\pi_1 (S^1\vee S^1)$, so the above works equally well.
But as I mentioned earlier, this was an exercise 1.1.17 in Hatcher, that is, this would be solved most appropriately without knowledge on e.g. the Seifert-Van Kampen theorem or covering spaces, which appear later in the textbook. So my question:
Is there a solution that only appeals to the basic definitions and lemmas on the fundamental groups, avoiding the use of the nontrivial theories like the Seifert-Van Kampen or covering spaces?