# Prove that any two maps are homotopic.

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?

• 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. – Brian Shin May 12 at 14:58
• @BrianShin Thanks, I didn't know that! – user665297 May 13 at 6:33

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 $$g_0$$.
• 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? – user665297 May 12 at 13:51