Freely homotopic maps $\phi_i:X\to Y$ induce the same map on $\pi_1$ if $\pi_1(Y,y_0)$is abelian 
Assume that $\phi_i: (X, x_0) \rightarrow (Y, y_0)$, for $i=0,1$ are freely homotopic.  Prove that if $\pi_1(Y,y_0)$ is abelian, then $\pi_1(\phi_0) = \pi_1(\phi_1).$

I know that this means that I have to prove that, for $f : I \rightarrow X$ as a closed loop about $x_0$, $[\phi_0 \circ f] = [\phi_1 \circ f]$, or $\phi_0 \circ f \simeq \phi_1 \circ f \text{ rel $\{0,1\}$}$.
But it would seem that $H(s,t) = \phi_t \circ f(s)$ satisfies the homotopy, as $H(s,0) = \phi_0 \circ f(s), H(s,1) = \phi_1 \circ f(s),\text{ and } H(0,t) = H(1,t) =  \phi_t(x_0) = \phi_0(x_0) = \phi_1(x_0)$ since each $\phi_i$ is a pointed map.
But I didn't use the fact that $\pi_1(Y,y_0)$ is abelian at all?  So what exactly did I miss here?
 A: For any $t$ define $\gamma_t:[0,1]\to Y, s\mapsto\gamma_t(s)=\phi_{s\cdot t}(x_0)$, this is a path that starts at $y_0$ and ends at $\phi_t(x_0)$. But this is actually also continuous in both $t$ and $s$.
Lets look at a path $\alpha$ in $(X,x_0)$, then:
$$\gamma_t^{-1}*(\phi_t\circ\alpha)*\gamma_t$$
will be (for fixed $t$) a path that starts at $y_0$, goes to $\phi_t(x_0)$, then goes through the loop $\phi_t\circ\alpha$ to end up at $\phi_t(x_0)$ again and then runs back to $y_0$.
This map is also continuous in $t$, so its actually a homotopy. Since for for any $t$ the path both starts and ends at $y_0$ it leaves the basepoint fix and is a homotopy of loops.
Now this map is a homotopy between what loops? It is a homotopy between
$$\gamma_0^{-1}*(\phi_0\circ\alpha)*\gamma_0\quad\text{   and }\quad \gamma_1^{-1}*(\phi_1\circ\alpha)*\gamma_1$$
so these two paths must have the same class in the fundamental group $\pi_1(Y,y_0)$. Note that $\gamma_0$ is the path that just stays at $x_0$, so the class of the first path is the same as $[\phi_1\circ\alpha ]=\pi_1(\phi_0)[\alpha]$. Note also that $\gamma_1$ starts at $y_0$ and ends at $y_0$, so its a loop. So the class of the second path is $[\gamma_1^{-1}]*\pi_1(\phi_1)[\alpha]*[\gamma_1]$.
If $\pi_1(Y,y_0)$ is commutative this second class is however the same as $\pi_1(\phi_1)[\alpha]$, so you get $\pi_1(\phi_0)=\pi_1(\phi_1)$ in this case.
