# Homotopy equivalence of paths in homotopy-equivalent spaces

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.

• 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? – freakish Oct 4 '17 at 9:58

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.
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.