# Fundametal group and continuous extension to $\Bbb R^n$

Let $$X ⊂ \Bbb R^n$$ be a non empty subset with $$n>0$$ and let $$x_0 ∈ X$$.

Let $$Y$$ be a non empty topological space and $$g : X → Y$$ a continuous map.

Suppose $$g$$ has a continuous extension defined on $$\Bbb R^n$$.

My question is: Could we affirm that the morphism $$g_∗ : π_1 (X, x_0 ) → π_1 (Y, g(x_0 ))$$ induced by $$g$$ is trivial?

I think that $$π_1 (\Bbb R^n, x_0 )$$ is trivial because $$\Bbb R^n$$ is simply connected, but we might find a subset $$X$$ that is not path connected in which case $$g_∗(π_1 (X, x_0 ))$$ is not trivial?

• It doesn't matter what subset of $X$ you find. A loop in $\pi_1(X,x_0)$ has to be able to pass through $\pi_1(\mathbb R^n, x_0) = 0$ to get to $g_*(\pi_1(X,x_0))$, which means you always pick up a nullhomotopy along the way. – Justin Young Feb 17 at 13:20
Let $$i: X\to \Bbb R^n$$ be the inclusion and $$g': \Bbb R^n \to Y$$ be the extension of $$g$$ such that $$g=g'\circ i$$.
We have $$g_*(\pi_1(X,x_0))=g'_*\circ i_*(\pi_1(X,x_0))=g'_*(0)=0$$.
Therefore $$g_*$$ is trivial.