Do freely homotopic maps induce the same homomorphism on fundamental groups? 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 $\sigma=H(x_0,-)$?
 A: 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 $g$. 
A: $\phi_\sigma$ 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 $\pi_1(Y,f(x_0))$. 
A: 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.
