Two basic problems about fundamental group and paths I have two problems that I don't know how to do )=.
i) Let X path connected. Let $x_0 , x_1 \in X$. Show that $\pi _1 \left( {X,x_0 } \right)$ is abelian iff for every $\alpha, \beta$ paths from $x_0$ to $x_1$ , we have $\widehat\alpha  = \widehat\beta$.
Where the hat denotes the homomorphism:
$$
\eqalign{
  & \widehat\alpha :\pi _1 \left( {X,x_0 } \right) \to \pi _1 \left( {X,x_1 } \right)  \cr 
  & \widehat\alpha \left( {\left[ f \right]} \right) = \left[ \alpha  \right]^{ - 1} \left[ f \right]\left[ \alpha  \right] \cr} 
$$
I did the side where the group is abelian, but the other I could not )=.
ii) The other is a property that I read in wikipedia. If $X$ and $Y$ are path connected, then $$
\pi _1 \left( {X\times Y} \right) \cong \pi _1 \left( {X\,} \right) \times \pi _1 \left( Y \right)
$$
I'm not sure if this nice property is easy to show. )=
 A: For $i$) "the other", I assume you're asking to show that if $\widehat{\alpha} = \widehat{\beta}$ for all $\alpha$ and $\beta$, then $\pi_1(X,x_0)$ is abelian.
Here's a hint:  Let $a,b\in \pi_1(X,x_0)$.  You want to show that $aba^{-1} = b$.  Is there a path $\alpha$ starting and ending at $x_0$ for which $\widehat{\alpha}(b) = aba^{-1}$?  Is there another easy path $\alpha'$ from $x_0$ to $x_0$ for which $\widehat{\alpha'}b = b$?
For $ii$), the point is that a map from $S^1$ into $X\times Y$ is nothing more than 2 maps, one from $S^1$ into $X$ and the other from $S^1$ into $Y$.
As a hint:  If $\pi$ and $\pi'$ are the projections from $X\times Y$ onto $X$ and $Y$, respectively, then show $[\gamma]\rightarrow(\pi_*[\gamma],\pi'_*[\gamma])$ is an isomorphism from $\pi_1(X\times Y)$ to $\pi_1(X)\times\pi_1(Y)$.
Edit  It seems there is a bit of confusion as to whether or not the correct assumption is "For all $x_0$ and $x_1$, for any two paths between them the induced maps are the same" or if it's "There exists two points $x_0$ and $x_1$ so that any 2 paths between them induce the same map."  I claim that the two assumptions are actually equivalent.
More specifically, in a path connected space $X$, if there is some pair of points $x_0$ and $x_1$ so that all paths between $x_0$ and $x_1$ induce the same map on $\pi_1$, then for all pairs of points $x$ and $y$, any path between them induces the same map on $\pi_1$.
Once and for all, pick a path $\gamma_1$ from $x_0$ to $x$ and a path $\gamma_2$ from $y$ to $x_1$.
Let $\alpha_1$ and $\alpha_2$ be two paths from $x$ to $y$.  Then the path $\alpha_I' = \gamma_1 * \alpha_i * \gamma_2$ is a path from $x_0$ to $x_1$.
Finally, let $f\in\pi_1(X,x)$.  Then $[\gamma_1 f \gamma_1^{-1}]\in \pi_1(X,x_0)$.  By assumption, $\widehat{\alpha_1'}[\gamma_1 f \gamma_1^{-1}] = \widehat{\alpha_2'}[\gamma_1 f \gamma_1^{-1}]$.
Cruching these two things and simplifying gives $$[\gamma_2^{-1} \alpha_1^{-1} f \alpha_1 \gamma_2] = [\gamma_2^{-1} \alpha_2^{-1} f \alpha_2 \gamma_2].$$  Applying $\widehat{\gamma_2^{-1}}$ to both sides shows that $\widehat{\alpha_1} = \widehat{\alpha_2}$ as claimed.
A: Following what Jason says, maybe you could start proving that you already have a bijection between sets of continuous maps 
$$
\Phi : \mathrm{Hom}(Z, X\times Y) \longrightarrow \mathrm{Hom} (Z, X) \times \mathrm{Hom}\mathrm(Z,Y)
$$ 
for any topological space $Z$, particularly for $Z = S^1$.
This is just the universal property of the product topology: a continuous map $f : Z \longrightarrow X \times Y$ is the same as a pair of maps $Z \longrightarrow X$ and $Z \longrightarrow Y$. Namely, $\Phi$ sends $f$ to $(p_Xf, p_Yf)$, where $p_X : X\times Y \longrightarrow X$ and $p_Y : X\times Y \longrightarrow Y$ are the natural projections. You should think who is the inverse $\Psi$ of this $\Phi$.
The next step then could be to show that (punctured) homotopy relationship preserves this bijection.
Namely, you should prove that if you have an homotopy $H: Z\times I \longrightarrow X\times Y$ between maps $f, g: Z \longrightarrow X\times Y$, then you also have homotopies between compositions: $p_Xf \sim p_X g$ and $p_Yf \sim p_Yg$.
Adding preserving base point maps and preserving base point homotopies at this stage should be harmless.
As a consequence, you will have a well-defined map between homotopy classes induced by $\Phi$:
$$
\widetilde{\Phi} : [(Z,z_0), (X\times Y,(x_0,y_0))] \longrightarrow [(Z,z_0), (X,x_0)] \times [(Z,z_0), (Y,y_0)] \ .
$$
Precisely,
$$
\widetilde{\Phi}([f]) = ([p_Xf], [p_Yf]) \ .
$$
(Here $x_0 \in X$, $y_0 \in Y$ and $z_0 \in Z$ are the base points.)
Finally, in order to get an inverse for this map between sets of (punctured) homotopy classes, you could repeat the same process for $\Psi$, obtaining some $\widetilde{\Psi}$ in the opposite direction.
A: The more general result on products is that if $\pi_1(X)$ denotes the fundamental groupoid of the space $X$, then for spaces $X,Y$, the natural morphism 
$$\pi_1(X \times Y) \to \pi_1(X) \times \pi_1(Y) $$
determined by the projections $X \times Y \to X, X \times Y \to Y$, is an isomorphism. ("Topology and groupoids" 6.4.4). 
