I have this exercise:
Let A be a subspace of $\mathbb{R}^n$; let $h \colon (A,a_0)\rightarrow (Y,y_0)$. Show that if h is extendabe to a continuous map of $\mathbb{R}^n$ into Y, then $h_*$ is the trivial homomorphism.
I found a solution online, it is very technical and I do not really understand it: http://dbfin.com/topology/munkres/chapter-9/section-52-the-fundamental-group/problem-5-solution/
I am wondering if this alternative solution is correct:
We must show that for any loop in A, based at $a_0$ we have that $[h\circ f]$ is a loop in Y, based at $y_0$ that is path-homotopic to the constant map at $y_0$.
When looking at the loop in Y: $I \rightarrow h \circ f$ it doesn't matter if we look at $f$ in $A$ or $\mathbb{R}^n$(this is the part I am unsure of?). So we look instead at it just as a loop in $\mathbb{R}^n$, but since this space is convex, this loop is nullhomotopic. Since h is extendable to $\mathbb{R}^n$ it makes sense to evaluate a loop in $\mathbb{R}^n$ by $h_*$. But since $[f]$ is trivial in $\mathbb{R}^n$, we get that $h_*([f])$ is a trivial loop, since $h_*$ is a group-homomorphism.
Is this correct? Are we allowed to argue like this, or do we need to to more work?, or is the argument wrong?
You have a pointed map $h:(A,a)\to (Y,y)$ with $A\subseteq \Bbb R^n$. Suppose this extends, that is, there is a map $\tilde h:(\Bbb R^n,a)\to (Y,y)$ such that $\tilde h i=h$ with $i:A\to\Bbb R^n$ the inclusion. Then, since $(-)_\ast$ is functorial, you have a factorisation $\tilde h_\ast i_\ast =h_\ast$. But $\Bbb R^n$ is contractible to the point $a$ by the homotopy $H:\Bbb R^n\times I\to\Bbb R^n$ such that $H(x,t)=tx+(1-t)a$. It follows that $\tilde h_\ast$ is trivial, and hence so is $h_\ast$. The idea is that if one can factor a map through a trivial map, the original map must be trivial too.