An inverse for a knowing isomorphism. > Let $i:X\rightarrow X\times Y$ and $j:Y \rightarrow X\times Y$ be maps defined by $i(x)=(x,y_0) $ and $j(y)=(x_0,y)$, where $x_0\in X$ and $y_0\in Y$. Prove that the mapping of $\pi(X,x_0)\times \pi(Y,y_0)$ into $\pi(X\times Y, (x_0,y_0))$ defined by $([\beta],[\gamma])\mapsto (i_*[\beta])(j_*[\gamma])$ is an isomorphism of the first group onto the second. I have a hint: prove itis the inverse of the isomorphism given by $\alpha\mapsto(p_*(\alpha),q_*(\alpha))$ where $p_*:X\times Y \rightarrow X$, $q_*:X\times Y\rightarrow Y$ are the projections onto the first and the second variable respectively. Name as $f$ and $g$ the first and the second isomorphism given respectively. Prove that $gf=1$ is easy, since $pi=1_{X}$ and $qj=1_{Y}$. But calculating the other one I got $$fg([\alpha])=[ip\alpha][jq\alpha] $$ and this should be equal to $[\alpha]$ but I get stuck. Somebody can show me the way? Thanks!

Write $\alpha(t) = (\alpha_1(t), \alpha_2(t))$. Then $ip\alpha(t) = (\alpha_1(t),y_0)$ and $jq\alpha(t) = (x_0, \alpha_2(t))$. That is, $ip\alpha = \alpha_1 \times c_{y_0}$ and $jq\alpha = c_{x_0} * \alpha_2$ (where $c_{x_0}$ and $c_{y_0}$ are the constant loops based at $x_0$ and $y_0$, respectively). Hence

$$ip\alpha * jq\alpha = (\alpha_1 \times c_{y_0}) * (c_{x_0} \times \alpha_2) \simeq (\alpha_1 * c_{x_0}) \times (c_{y_0} * \alpha_2) \simeq \alpha_1 \times \alpha_2 = \alpha,$$

and consequently

$$[ip\alpha][jq\alpha] = [ip\alpha * jq\alpha] = [\alpha].$$