I have the following problem. Let $X$ be a subspace of $\mathbb{R}^n$ and $Y$ some topological space. Also, let $\psi: (X,x_0) \to (Y,y_0)$ be a continuous map. If there exists a continuous extension of $\psi$, say $\tilde{\psi}: \mathbb{R}^n \to Y$, then $\psi$ induces a trivial homomorphism in the fundamental groups (I guess of $X$ and $Y$)?
Not only I cannot show the above but also I am not sure I understand the question..
Let $X$ be a subspace of $\mathbb R^n$ and let $Y$ be some topological space.
Consider the inclusion function $j : X \to \mathbb R^n$ where $x \mapsto x$.
Suppose you have $\varphi : (X, x_0) \to (Y, y_0)$ a continuous map that extends to $\mathbb R^n$, that is, there exists $k : (\mathbb R^n, x_0) \to (Y, y_0)$ such that $k \vert_X = \varphi$. But recall that $k \vert_X = k \circ j = \varphi$ since we have $$(X, x_0) \xrightarrow{j} (\mathbb R^n, x_0) \xrightarrow{k} (Y, y_0)$$ which induces $$\pi_1(X, x_0) \xrightarrow{j_*} \pi_1(\mathbb R^n, x_0) \xrightarrow{k_*} \pi_1(Y, y_0).$$
Notice that $\pi_1(\mathbb R^n, x_0) = 0$ since $R^n$ is convex and thus every loop contracts to a point via the straight line homotopy, so $k_*$ is trivial.
We want to show that $\varphi_* : \pi_1(X, x_0) \to \pi_1(Y, y_0)$ is trivial.
But since $\varphi = k \circ j$, then $\varphi_* = k_* \circ j_*$.
Take a loop $f$ based at $x_0$ in $X$ and consider $[f] \in \pi_1(X, x_0)$. Observe that $\varphi_*\big([f]\big) = k_*\big(j\big([f]\big)\big) = k_*\big([f]\big) = 0$.
Conclude that $\varphi_*$ is trivial.