Nonhomotopic retractions $S^1\vee S^1\to S^1$ without Seifert-Van Kampen?

Question

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?

Answer 1

Recall that a map $f:A\vee B\rightarrow X$ with domain a wedge of spaces is completely determined by its restrictions to each of the wedge summands.

Therefore, for each $k\in\mathbb{Z}$ let $\underline k:S^1\rightarrow S^1$ be a chosen map of degree $k$. Consider the wedge as the subspace $S^1\vee S^1=(S^1\times \ast)\cup (\ast\times S^1)\subseteq S^1\times S^1$ of pairs of points such that at least one point is the basepoint $\ast$ of $S^1$ and define

$$r_k':S^1\vee S^1\rightarrow S^1$$

by

$$r_k'|_{S^1\times\ast}=pr_1,\qquad r_k'|_{\ast\times S^1}=\underline k\circ pr_2.$$

Finally set $r_k=in_1\circ r'_k:S^1\vee S^1\rightarrow S^1\vee S^1$, where $in_1:S^1\cong S^1\times\ast\hookrightarrow S^1\vee S^1$ is the inclusion into the first factor.

Then for each $k$, the map $r_k$ is a retraction of $S^1\vee S^1$ onto the subspace $S^1\times\ast$, since both components

$$r_k|_{S^1\times\ast}=in_1\circ pr_1|_{S^1\times\ast}=in_1,\qquad r_k|_{\ast\times S^1}=in_1\circ\underline k\circ pr_2|_{\ast\times S^1}$$

take values in the subspace $S^1\times\ast \subseteq S^1\vee S^1$, and in particular, the first component is just the inclusion of this subspace. Note, however, that the second component has degree $k$.

We claim that $r_k\not\simeq r_l$ for $k\neq l$. Assume to the contrary, that for integers $k\neq l$ there exists a homotopy $F:r_k\sim r_l$. Then the map

$$pr_2\circ F\circ (in_1\times I):S^1\times I\rightarrow S^1$$

is a homotopy $\underline k\sim\underline l$ between a map of degree $k$ and a map of degree $l$. Assuming the knowledge that

$$\pi_1S^1\cong\mathbb{Z},$$

with its elements indexed by degree, we see that for $k\neq l$, such a homotopy cannot exist. Therefore the homotopy $F$ cannot exist, and it must be that $r_k\not\simeq r_l$.

Answer 2

Just following the idea of Tyrone I write the proof with less notation :---- Note that, $\Bbb S^1\lor\Bbb S^1=\big(\Bbb S^1\times\{1\}\big)\bigcup\big(\{1\}\times\Bbb S^1\big)$. For $k\in \Bbb N$ define retraction $r_k:\Bbb S^1\lor\Bbb S^1\to\Bbb S^1\times\{1\}$ by $$r_k\big(a,1\big)=(a,1),\forall a\in \Bbb S^1,$$$$r_k(1,b)=(b^k,1),\forall b\in \Bbb S^1.$$ Hence, $r_k\big|_{\Bbb S^1\times\{1\}}=\text{Id}_{\Bbb S^1\times\{1\}}$ and $r_k\big|_{\{1\}\times\Bbb S^1}:\{1\}\times\Bbb S^1\to \Bbb S^1\times\{1\}$ is $k$-fold covering map. So $r_k\not\cong r_l$ if $k\not=l$. Actually, if $p:(X,x_0)\to (Y,y_0)$ is a $n$-fold covering then $p_*\big(\pi_1(X,x_0)\big)$ is a subgroup of $\pi_1(Y,y_0)$ of index $n$.