Show that if $h$ is extendable to a continuous map of $\Bbb R^n$ into $Y$, then $h_*$ is the trivial homomorphism.

Question

Let $A$ be a subspace of $\Bbb R^n$; let $h:(A,a_0) \to (Y,y_0)$. Show that if $h$ is extendable to a continuous map of $\Bbb R^n$ into $Y$, then $h_*$ is the trivial homomorphism.

I can't get any clue how to deal with it...

Answer 1

Let $i: A \longrightarrow \mathbb{R}^n$ be the inclusion.

Extend $h$ to a continuous function $k:\mathbb{R}^n \longrightarrow Y$.

Then $h = k \circ i$. So $h^* = k^* \circ i^*$ because the fundamental group is functorial.

But $k^*$ is the trivial homomorphism because its domain is $\pi_1(\mathbb{R}^n, a_0) = \{ e \}$ which is the trivial group. So $h^* = k^* \circ i^*$ is trivial.