# Prove that $(f_0,g_0)\simeq(f_1,g_1)$ rel $\{0,1\}.$

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.

$$\require{AMScd}$$

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

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 $$X \times Y$$: $$\begin{matrix} && I \times I && \\ & {\scriptsize G}\swarrow & ~\downarrow{\scriptsize K} & \searrow{\scriptsize H} & \\ X & \underset{\pi_1}{\leftarrow} & X \times Y & \underset{\pi_2}{\rightarrow} & Y \end{matrix}$$
• Thank you. To preserve the order I think it should be $K=(H,G)$. Are we using $\pi_i$ to find $K$? Aug 12, 2019 at 15:05