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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?

Thanks for your help.

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    $\begingroup$ 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. $\endgroup$ – Justin Young Feb 17 at 13:20
  • $\begingroup$ Thank you, it makes sense now $\endgroup$ – PerelMan Feb 17 at 13:30
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Based on Justin's comment:

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.

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