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I was trying to show the two following things: 1) any two maps from a topological space $X$ to $I=[0,1]$ are homotopic, 2) any two maps from $I=[0,1]$ to a path connected set $Y$ are homotopic.

For the second I thought I could do it like this:

Let $f:I\rightarrow Y$ en $g:I\rightarrow Y$ continuous maps. Now define a function $F:I\times I\rightarrow Y$ such that $F(x,t)=(1-t)f((1-t) x) + t g(tx)$. Then we see that $F(x,0)=f((1-0) x) + 0\cdot g(0\cdot x)=f(x)$ and $F(x,1)=(1-1)f((1-1) x)+1\cdot g(1\cdot x)=f(x)$. Furthermore, $F$ is continuous since it's a composition of continuous functions. Now we can conclude that $f$ and $g$ are homotopic.

However, I don't use the fact that $Y$ is path connected, so I think this can't be right. I already found this, however, I don't understand what the $e_{f(0)}$ and $e_{g(0)}$ are. Furthermore, I found this, but in the answers they don't create the $F$-function.

Could someone help me find the right $F$-function to show this? And also for the first part?

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    $\begingroup$ The reason the function $F$ you have written down does not work is because you use addition and multiplication. For a space $Y$ and two points $a,b\in Y$, what does $a+b$ mean? Unless $Y$ has a group structure, it means nothing. In particular, the expression you have written down doesn't quite make sense. $\endgroup$ – Brian Shin May 12 at 14:58
  • $\begingroup$ @BrianShin Thanks, I didn't know that! $\endgroup$ – user665297 May 13 at 6:33
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For the first question it is enough to show that the map $f$ is homotopic to the constant map $g(x)=0$, define $H_t(x)=tf(x)$.

For the second, you can also show that any map from $I\rightarrow Y$ is homotopic to $g(t)=y_0, y_0\in Y$. Firstly $f$ is homotopic to $g_0(x)=f(0)$ by using $H_t(x)=f(tx)$, consider a path $c:[0,1]\rightarrow Y$ such that $c(0)=f(0), c(1)=y_0$, write $H'_t(x)=c(t)$, $H'$ defines an homotopy between $g$ and $g_0$.

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  • $\begingroup$ Thanks! You said 'it is not to show' but you meant 'it is to show' right? In that case I understand that part. For the second, you have $f\simeq g_0$ and $g_0\simeq g$, can we then conclude $f\simeq g$ by the fact that homotopy is an equivalence relation? $\endgroup$ – user665297 May 12 at 13:51

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