0
$\begingroup$

I know that:

  • If a space X deformation retracts onto a subset A then they are homotopy equivalent.
  • If a path $f_0:I\rightarrow X$ can be continuously deformed to $f_1: I \rightarrow X$ then $f_0$ and $f_1$ are homotopy equivalent paths & this defines an equivalence relation between paths.

If I fix endpoints in X, is the map from a homotopy class of paths with these endpoints in X to a homotopy class of paths in A well-defined? Should I be able to apply a homotopy mapping X to itself, preserving endpoints of this segment, and then have this homotopy induce a homotopy on any path contained in X, and receive the same homotopy if I apply the retraction before or after I follow the path?

I feel like it is, but I'm not sure how to justify it or what theorem I can use to justify it. Working in Hatcher, chapter 1.

$\endgroup$
  • $\begingroup$ Every homotopy that preserves endpoints is a (general) homotopy. So the answer to the first question is: yes. As for the other question I don't understand it. Perhaps write in symbols what you you are trying to do? $\endgroup$ – freakish Oct 4 '17 at 9:58
0
$\begingroup$

Let $X, Y$ topological space and $f: X \to Y$ a continuos map. Let $x_0, x_1 \in X$ points (not necessarily different) and $y_0 = f(x_0), y_1 = f(x_1) \in Y$. Let $C(X; x_0, x_1)$ the set of paths in $X$ with initial point $x_0$ and end point $x_1$ and $C(Y; y_0, y_1)$ a similar set, let $\sim$ the equivalence relation you mentioned ( existence of a homotopy). We want to define a map $$f_*: C(X; x_0, x_1)/\sim \, \to \, C(Y; y_0, y_1)/\sim$$ Ok, let $\gamma: [0,1] \to X$ a path with $\gamma(0) = x_0$ and $\gamma(1) = x_1$, i.e., $\gamma \in C(X, x_0, x_1)$, define $f\gamma: [0,1] \to Y$ with $(f\gamma)(t) = f(\gamma(t))$. We have $f\gamma \in C(Y; y_0, y_1)$. Take $$f_*([\gamma])=[f \gamma]$$ Is it well defined? Let $\gamma' \in C(X, x_0, x_1)$ other path with $\gamma \sim \gamma'$, i.e., exist a homotopy $$H:[0,1] \times [0,1] \to X$$ with $$H(0,s) = x_0 \quad\quad H(1, s) = x_1\\H(t, 0) = \gamma(t) \quad \quad H(t, 1) = \gamma'(1)$$ Take $$fH: [0,1]\times[0,1] \to Y: (t,s) \mapsto fH(s,t) = f(H(s,t))$$ Note that $$Hf(0,s) = f(H(0,s)) = f(x_0) = y_0\\Hf(1,s) = f(H(1,s)) = f(x_1) = y_1\\Hf(t,0) = f(H(t,0)) = f(\gamma(t)) = (f \gamma)(t)\\Hf(t,1) = f(H(t,1)) = f(\gamma'(t)) = (f \gamma')(t)$$ So $fH$ is a homotopy between $f\gamma$ and $f \gamma'$. Then $f \gamma \sim f \gamma'$ and $f_*$ is well defined.

Is that what you want? I'm not sure what you mean in your second question.

You said nothing about concatenation/product of paths, but this map preserve concatenations in the follow sense: if $\gamma_1 \in C(X; x_0, x_1)$ and $\gamma_2 \in C(X; x_1, x_2)$ then $$f_*([\gamma_1] \cdot [\gamma_2]) = f_*([\gamma_1]) \cdot f_*([\gamma_2])$$.

Now, if $Y=X$ and $y_0 = x_0, y_1 = x_1$, we have $$f_*: C(X; x_0, x_1)/\sim \, \to \, C(X; x_0, x_1)/\sim \,: [\gamma] \mapsto f_*([\gamma]) = [f \gamma] $$.

A more interesting case result if $x_0 = x_1=x$, $y_0 = y_1=y$. Then, we defined $$f_* : \pi_1(X, x) \to \pi_1(Y, y)$$ As we observed, it's a group homomorphism and we can proof that $$\pi_1: \mathbf{Top*} \to \mathbf{Grp}\\(X,x) \mapsto \pi_1(X,x)\\f \mapsto f_*$$ is a (covariant) functor.

See Lima, Elon - Fundamental Group and Covering Space for a discution about free homotopy and this kind of stuff.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.