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

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?

• I think you got it right, that's also the example I thought of. What's important is that you think of the homotopy as a homotopy of retractions, i.e any slice of the homotopy is also a retraction. With this in mind, it shouldn't be hard to prove, because you would sort of get a homotopy between loops that go different number of times around the second circle ...
– Laz
Commented Nov 7, 2018 at 2:11
• @Laz Wait, should the statement of the exercise be read as "there is no homotopy of retractions between two retractions", not that they are not homotopic just as maps between the spaces? I'm not a native speaker of English, but it feels like there's a jump from the notion of homotopy of maps to that of retractions. Commented Nov 7, 2018 at 2:17
• The exercise never states that the homotopy should be a homotopy of retractions, I just observe that in Algebraic Topology, when you have a map with a certain property (e.g it sends a subspace $A$ to a subspace $B$, as in the case of loops) it is customary that when you talk about a homotopy of it, it's a homotopy with the same property. You can try both problems. The more general version should be a little more difficult, but I think it's also doable. I sincerely don't know which one did Hatcher want the reader to solve.
– Laz
Commented Nov 7, 2018 at 2:30
• Odd, I didn't know that Hatcher defined retractions to be maps $X\to X$. It makes it more awkward to state correctly the definition than the definition as maps $X\to A$, which makes more sense categroy-theoretically (and so for more general settings for homotopy theory) Commented Nov 7, 2018 at 6:40
• @Max No topologists define retraction as a map between different spaces; you can find the definition of retractions on Hatcher, pg 3, for instance. Commented Nov 7, 2018 at 10:09

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 $$r_k\not\simeq r_l$$.

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 $$n$$.