Prove $j_{*}:\pi_{1}(X,b) \rightarrow \pi_{1}(Y,b)$ is surjective for certain X,Y.

Question

Let $P=(2,0)$ and $O=(0,0)$. Let $Y=\mathbb{R}^2\backslash\{O\}$ and $X = \mathbb{R}^{2}\backslash \{O,P\}$ and let $j:X \rightarrow Y$ be an inclusion.

Prove $j_{*}:\pi_{1}(X,b) \rightarrow \pi_{1}(Y,b)$ is surjective.

I know that because the fundamental groups of these spaces are not the same so $j$ can't be injective So i only need to prove the surjection.

So far i've tried to do this:

i need to find for all $[g] \in \pi_{1}(Y,b)$ an $f \in \pi_{1}(X,b)$ such that $[j \circ f]$ = [g]

Because $j$ is an inclusion we can choose $f=g$ for loops that don't contain the point $P$, but the main problem is how to choose $f$ if $g$ contains the point $P$?

Any tips/hints/definitions/lemma's which can help me are much appreciated

With kind regards,

Kees Til

Answer 1

The answer to this question gives the exact method that you will need to prove surjectivity of $j_*$.