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Homotopy

hmmm16

Member
Feb 25, 2012
31
I am just starting to do some algebraic topology (not really a strong point of mine) and I am struggling to get a good grasp of "homotopic relatove to a set $A$. The definition of homotopic that I am using is (I think it's standard, but just in case and so that my notation is fine) :

Two continuous maps $f_0,f_1$ are homotopic if there is a continuous map $F:X\times I\rightarrow Y$ such that $F(x,0)=f_0(x)$ and $F(x,1)=f_1(x)$



So I understand that intuitively two continuous maps $f_0,f_1:X\rightarrow Y$ are homotopic if there is a "continuous translation of one to the other".

However I am bit confused as to how to think of "homotopic relative to a set $A$". To be precise that is:


If there is a homotopy $F:X\times I\rightarrow Y$ between $f_0$ and $f_1$ such that $F(a,t)=f_0(a)$ for all $a\in A$ and all $t\in I$

I'm not really understanding what this definition is telling me?

Thanks for any help
 

Turgul

Member
Jan 13, 2013
35
Typically a homotopy $F:X \times I \rightarrow Y$ need only be continuous, start at $f_0$ and end up at $f_1$. This condition of being a homotopy relative to $A \subseteq X$, as you have defined it, is exactly the statement that the homotopy must be constant on $A$.

A trivial example is the following: if $f_0=f_1=F(-,t)$ for all $t \in I$, then $F$ is a homotopy of $f_0$ to $f_1$ relative to any (and every) subset of $X$, since the homotopy does nothing to the function at all!

Consider the maps $f_0: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $v \mapsto v$ (the identity) and $f_1$ defined by $v \mapsto 0$ (the constant zero map). There is a homotopy $F$ from $f_0$ to $f_1$ defined by $(v,t) \mapsto (1-t)v$. Note that
\[f_0(0) = f_1(0) = F(0,t) = 0\]
for any $t \in [0,1]$ but
\[f_0(v) \neq f_1(v) \neq F(v,t)\]
for any $v \neq 0$ and $t \in (0,1)$. This is an example of a homotopy from $f_0$ to $f_1$ relative to $\{0\} \subset \mathbb{R}^2$.