Homotopy classes of self-maps on $\mathbb{S}^1\vee\mathbb{S}^1$

Consider the two inclusions $$\eta_i:\mathbb{S}^1\to \mathbb{S}^1\vee\mathbb{S}^1$$. I claim that the following map is injective $$(\eta_1\sqcup\eta_2)^*:[\mathbb{S}^1\vee\mathbb{S}^1,\mathbb{S}^1\vee\mathbb{S}^1]\to [\mathbb{S}^1\sqcup\mathbb{S}^1,\mathbb{S}^1\vee\mathbb{S}^1].$$ In other words, two self-maps $$f,g:\mathbb{S}^1\vee\mathbb{S}^1\to \mathbb{S}^1\vee\mathbb{S}^1$$ are homotopic if for the restrictions $$f_i:=f\circ\eta_i$$ and $$g_i:=g\circ \eta_i$$, we have $$f_i\simeq g_i$$.

If we would work with based homotopy, the statement would be clear. However, the homotopies $$f_1\simeq g_1$$ and $$f_2\simeq g_2$$ may move the base point …

• what is the question? Can you clearly indicate somewhere? Commented Apr 23, 2019 at 23:09
• I am wondering whether my claim is true or not. Commented Apr 23, 2019 at 23:18
• Do your square brackets denote based or unbased homotopy classes? Commented Apr 24, 2019 at 12:14
• Unbased homotopy classes – which seems to be the problem. Commented Apr 24, 2019 at 12:57

Let $$f: S^1 \vee S^1 \rightarrow S^1 \vee S^1$$ be $$l \vee r$$ where $$l$$ means wrap around the left copy of $$S^1$$ and $$r$$ the right, and $$g:S^1 \vee S^1 \rightarrow S^1 \vee S^1$$ be $$l \vee (lr)^{-1}r(lr)$$. The image of $$[f]$$ and $$[g]$$ under the map is the same since $$[l]=[l]$$ and $$[(lr)^{-1}r(lr)]=[r]$$ (since one is the conjugate of the other, see Hatcher).
Now $$f$$ is not homotopic to $$g$$ since if they were we could look at the path of the basepoint under such a homotopy. It suffices to show this path, really a loop, does not commute with $$[l]:S^1 \rightarrow S^1 \vee S^1$$, since then conjugation by this loop changes the basepointed homotopy type of $$l$$, meaning that there is no homotopy starting at $$l$$, ending at $$l$$, and taking the basepoint along such a path (see Hatcher for the relation between conjugation and homotopy).
A homotopy from $$r$$ to $$(lr)^{-1}r(lr)$$ necessarily has the homotopy class of the loop of the basepoint equal to a word with $$r$$ in it. This is because we are conjugating $$r$$ by a word to get $$(lr)^{-1}r(lr)$$ (again, see Hatcher for this relation). Such a thing necessarily does not commute with $$l$$, so there is no homotopy from $$l \vee r$$ to $$l \vee (lr)^{-1}r(lr)$$.
• Thank you! Could you explain a bit more why it is sufficient to show that the loop described by the basepoint does not commute with $[l]$? I understood that $[\mathbb{S}^1,Y]$ is $\pi(Y)/\text{conjugation}$ and the above fact would give us that $[l]$ and $[*]\cdot [l]\cdot [*]^{-1}$ are different in $[\mathbb{S}^1,Y]$ where $[*]$ is the loop of the basepoint. But how does this help us? Commented Apr 24, 2019 at 13:01
• What we want to show is that any homotopy starting at $l \vee (lr)^{-1}r(lr)$ which on the right circle ends at $r$ cannot end at $l$ on the left circle. Since the homotopy class of the map at the end of the homotopy on the left circle is what you get when you conjugate $l$ by the loop the basepoint takes, if we show that the loop does not commute with the map of the left circle into the space, then it has to result in a different basepointed homotopy class. Since any map is homotopic to itself, the resulting map cannot be $l$. Commented Apr 24, 2019 at 15:06