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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 $\sigma=H(x_0,-)$?

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The title is a bit misleading; it seems to be about a more elementary question than the body. –  Qiaochu Yuan Jun 30 '14 at 18:32
@QiaochuYuan Why is that? Should I put "freely homotopic" in the title? –  Christoph Jun 30 '14 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 '14 at 19:58
@QiaochuYuan Alright, I changed that. Thanks! –  Christoph Jun 30 '14 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 $g$.

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I'm not quite sure how this gives two freely homotopic maps $Y\to X$ inducing different homomorphisms? –  Christoph Jun 30 '14 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 '14 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 '14 at 19:32
If I consider $X=S^1$ and $Y$ the bouquet of two circles, then the natural candidates for $f$ and $g$ are the inclusions of $S^1$ into each circle. They induce different homomorphism, but I don't see how $f$ and $g$ are homotopic. –  Chilote Apr 5 at 1:40
@Chilote: they aren't. The condition for $f$ and $g$ to be (freely) homotopic is that, as elements of $F_2$, they are conjugate, which the two generators of $F_2$ are not. Being conjugate implies, in particular, mapping to the same element of the abelianization; said more topologically, a necessary condition for $f$ and $g$ to be (freely) homotopic is that they induce the same map on $H_1$. –  Qiaochu Yuan Apr 5 at 2:24

$\phi_\sigma$ 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 $\pi_1(Y,f(x_0))$.

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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 '14 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.

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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 '14 at 18:48

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