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#### Olinguito

##### Well-known member

- Apr 22, 2018

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This is a problem I thought of myself, and I’d like to be sure that my solution is all right. Let $X,Y,Z$ be topological spaces, with continuous functions $f_1,f_2:X\to Y$ and $g_1,g_2:Y\to Z$. I wish to show that the compositions $g_1\circ f_1:X\to Z$ and $g_2\circ f_2:X\to Z$ are homotopic.

Now $f_1,f_2$ homotopic $\implies$ there is a homotopy $H_f:X\times I\to Y$ (where $I$ is the closed interval $[0,\,1]$) with $H_f(x,0)=f_1(x)$ and $H_f(x,1)=f_2(x)$ for all $x\in X$.

Similarly there is a homotopy $H_g:Y\times I\to Z$ with $H_g(y,0)=g_1(y)$ and $H_g(y,1)=g_2(y)$ for all $y\in Y$.

Then I define $H:X\times I\to Z$ as follows:

$$H(x,t)\ =\ \begin{cases}g_1\left(H_f(x,2t)\right) & 0\le t\le\frac12 \\\\ H_g\left(f_2(x), 2t-1\right) & \frac12\le t\le1\end{cases}$$

for all $x\in X$. The idea is to first deform $g_1\circ f_1$ to $g_1\circ f_2$, and then the latter to $g_2\circ f_2$. Check:

- At $t=0$, $H(x,0)=g_1\left(H_f(x,0)\right)=g_1(f_1(x))$;

. - At $t=1$, $H(x,1)=H_g\left(f_2(x),1)\right)=g_2(f_2(x))$;

Hence $g_1\circ f_1\simeq g_2\circ f_2$.

Is my solution okay?