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Let $f_0\simeq f_1$ rel $\{0,1\}$ and $g_0\simeq g_1$ rel $\{0,1\}$ paths in $X,Y,$ respectively. If $(f_i,g_i)$ is the path in $X\times Y$ defined by $t\mapsto(f_i(t),g_i(t))$ for $i=0,1$ prove that $(f_0,g_0)\simeq(f_1,g_1)$ rel $\{0,1\}.$

By hypothesis, there is an homotopy $H:f_0\simeq f_1$ rel $\{0,1\}$, $H:I\times I\to X$ and there is an homotopy $G:g_0\simeq g_1$ rel $\{0,1\}$, $G:I\times I\to Y$.

Notation $F_i=(f_i,g_i):I\to X\times Y.$

We want $K:(f_0,g_0)\simeq(f_1,g_1)$ rel $\{0,1\}, K:I\times I\to X\times Y.$

How could I find the homotopy?

I have tried to draw the diagram but I do not know how to link accordingly.

If someone could help me, thank you.

$\require{AMScd}$

\begin{CD} X @<<{\text{H}}< I\times I @>{\text{G}}>> Y \\ @. @VVKV \\ @. X\times Y @<{\text{F0}}<{\text{F1}}< I \end{CD}

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You might be overthinking it. The map you want is $K = (G, H) : I \times I \to X \times Y$; explicitly, $K(s,t) = (G(s,t), H(s,t))$ for all $(s,t) \in I \times I$.

The corresponding diagram is the usual diagram representing the universal property of the product $X \times Y$: $$ \begin{matrix} && I \times I && \\ & {\scriptsize G}\swarrow & ~\downarrow{\scriptsize K} & \searrow{\scriptsize H} & \\ X & \underset{\pi_1}{\leftarrow} & X \times Y & \underset{\pi_2}{\rightarrow} & Y \end{matrix} $$

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  • $\begingroup$ Thank you. To preserve the order I think it should be $K=(H,G)$. Are we using $\pi_i$ to find $K$? $\endgroup$ Aug 12, 2019 at 15:05
  • $\begingroup$ @vino: That's right, my bad. When I was in school we learnt that G comes before H ;) $\endgroup$ Aug 12, 2019 at 17:42

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