# Homotopy of composition of homotopic functions

#### Olinguito

##### Well-known member
Hi all. 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 $t=1$, $H(x,1)=H_g\left(f_2(x),1)\right)=g_2(f_2(x))$;
and at the halfway stage $t=\frac12$, $g_1\left(H_f(x,1)\right)=g_1(f_2(x))$ and $H_g\left(f_2(x),0)\right)=g_1(f_2(x))$, so $H\left(x,\frac12\right)$ is well defined thereat.

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

Is my solution okay? #### Euge

##### MHB Global Moderator
Staff member
Hi, Olinguito !

In the problem statement, it is not stated that $f_1$ is homotopic to $f_2$ and $g_1$ is homotopic to $g_2$, which you used in your argument. I assume, however, that this was an intended assumption, in which case your proof is correct. Here is another proof: using the same $H_g$ and $H_f$ that you used, set $H = H_g \circ (H_f \times \text{id})$, i.e., $H(x,t) = H_g(H_f(x,t),t)$. Then $H$ is continuous with $H(x,0) = H_g(f_1(x),0) = g_1(f_1(x))$ and $H(x,1) = H_g(f_2(x),1) = g_2(f_2(x))$. Thus, $H$ is homotopy from $g_1 \circ f_1$ to $g_2 \circ f_2$.