Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f,g\colon X\to Y$ be two continuous maps that are freely homotopic, such that there is some $x_0\in X$ with $f(x_0)=g(x_0)$. Is it true that the induced homomorphisms $f_*,g_*\colon \pi_1(X,x_0)\to\pi_1(Y,f(x_0))$ are equal?

When $f,g$ are pointed homotopic relative $x_0$, the statement holds. When we only have a free homotopy $H$ with $H(-,0)=f, H(-,1)=g$, we get $f_*=\phi_\sigma\circ g_*$, where $\sigma=H(x_0,-)$ is the path travelled by $f(x_0)$ during the homotopy $H$ and $\phi_\sigma$ is the automorphism of $\pi_1(Y,f(x_0))$ given by $[\omega]\mapsto[\sigma\omega\overline{\sigma}]$. Thus the problem comes down to the question if $\phi_\sigma=\operatorname{id}$ for the loop $\sigma=H(x_0,-)$?

share|improve this question
The title is a bit misleading; it seems to be about a more elementary question than the body. –  Qiaochu Yuan Jun 30 at 18:32
@QiaochuYuan Why is that? Should I put "freely homotopic" in the title? –  Christoph Jun 30 at 18:56
yes. Otherwise it sounds like you're asking about based homotopy, since that's the sort of thing that would be an exercise in introductory algebraic topology. –  Qiaochu Yuan Jun 30 at 19:58
@QiaochuYuan Alright, I changed that. Thanks! –  Christoph Jun 30 at 20:51

3 Answers 3

up vote 5 down vote accepted

No. Let $X = S^1$ (note that if there is a counterexample then composing with a loop that distinguishes the induced maps on fundamental groups shows that there is a counterexample where $X = S^1$) and let $Y$ be any space with nonabelian fundamental group. Pick a loop $h \in \pi_1(Y)$ such that there exists $g \in \pi_1(Y)$ with $ghg^{-1} \neq h$, and consider the free homotopy from $h$ to $ghg^{-1}$ given by transporting the loop around $g$.

share|improve this answer
I'm not quite sure how this gives two freely homotopic maps $Y\to X$ inducing different homomorphisms? –  Christoph Jun 30 at 18:49
@Christoph Look at where the generator of $\pi_1(S^1)$ maps. In general, there is a bijection between conjugacy classes in $\pi_1(Y)$ and free homotopy classes of maps $S^1 \rightarrow Y$. In any space with non-abelian fundamental group, then, there are two freely homotopic maps $S^1 \rightarrow Y$ that are not basepoint-homotopic; each map sends the generator of $\pi_1(S^1)$ to a different element in $\pi_1(Y)$. –  Mike Miller Jun 30 at 19:21
Oh I get it, the generator of $\pi_1(S^1)$ is represented by $\operatorname{id}_{S^1}$, so the images are just $[h]$ and $[ghg^{-1}]$ itself, which aren't equal. –  Christoph Jun 30 at 19:32

$\phi_\sigma$ is an inner automorphism of the group $\pi_1(Y,f(x_0))$, namely conjugation by $[\sigma]$. So $f_*=g_*$ if and only if $[\sigma]$ is in the center of the group $\pi_1(Y,f(x_0))$.

share|improve this answer
Incidentally, this shows that the induced homomorphisms on fundamental groups are homotopic (where, by definition, homotopic means related by conjugation as above). There is a category which deserves to be called the homotopy category of groups whose objects are groups and whose morphisms are homotopy / conjugacy classes of homomorphisms; it models the homotopy category of path-connected, but not necessarily pointed, homotopy $1$-types. –  Qiaochu Yuan Jun 30 at 20:00

Let $h :X \times I \to Y$ be a (possibly non-pointed) homotopy $h:f\simeq g$, you get a path $a(t):=H(x,t)$ from $f(x)$ to $g(x)$. Conjugation under $a$ defines an isomorphism $\gamma[a]:\pi_1(Y,f(x)) \to \pi_1(Y,g(x))$, by means of $[f]\mapsto [a*f*a^{-1}]$. Now, what is true is that you get: $$\gamma[a]\circ f_*=g_*$$ Hence the two maps coincide up to a (non canonical) isomorphism.

share|improve this answer
Yeah that's what I figured in my question already. That's why I asked if this isomorphism was trivial for some reason. –  Christoph Jun 30 at 18:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.