I recently happened to come across this interesting result in topology:
Let $A$ be a subspace of $\mathbb{R}^n$ and let $h:(A,a_0) \rightarrow (Y,y_0)$. Then, if $h$ is extendable to a continuous map of $\mathbb{R}^n$ into $Y$, then the induced homomorphism $h_*: \pi_1(A,a_0) \rightarrow \pi_1(Y,y_0)$ is the trivial homomorphism (the homomorphism that maps everything to the identity element).
Remark: In the above result, $h:(A,a_0) \rightarrow (Y,y_0)$ denotes the continuous map $h:A \rightarrow Y$ that carries the point $a_0$ of $A$ to the point $y_0$ of $Y$. That is, $h(a_0)=y_0$.
My question is: Is the converse of the above statement true? That is, if $h_*$ is the the trivial homomorphism, then can $h:A \rightarrow Y $ be extended to a continuous map of $\mathbb{R}^n$ into $Y$?
I have been trying to answer this question for a while, but to no avail. So, any help/hint will be extremely useful.
Also, as a side note, I'm new to Mathematics Stack Exchange and this is my very first question. So, I'm still in the process of getting used to the order of things here, and any suggestion for improvement when it comes to posting questions will be appreciated. Thanks.
Welcome to MSE!
No. :P
As a simple example, take $S^2 \subseteq \mathbb{R}^3$, and look at the identity map $S^2 \to S^2$. Everything in sight is simply connected, so every fundamental group is trivial and the obstruction cannot live there. However we know there is no continuous $\mathbb{R}^3 \to S^2$ extending the identity (here we work with an analogous argument using $\pi_2$ instead of $\pi_1$. We know $\pi_2 S^2 = \mathbb{Z}$ and $\pi_2 \mathbb{R}^3 = 1$. Of course, homology groups or cohomology groups would also do the trick, and are more computationally friendly).
The next obvious question is "if all of the $\pi_n$s vanish, does the map extend?". This (as well as the related question for cohomology groups, etc.) is one of the founding questions in obstruction theory. See chapter $7$ of this set of lecture notes for more information, or this paper of Steenrod for a (much more readable) account.
I hope this helps ^_^