Prove $\iota ': \pi _1(A,a) \to \pi _1(B,a)$ is a bijection, if $C\setminus \{b\}\subseteq B$ is simply connected.

Question

I am having trouble solving this problem:

Let $A\subseteq B$ and consider the inclusion $\iota : (A, a) \to (B, a)$ with pointed spaces $(A, a)$, $(B, a)$. Let $A=B \setminus \{b\}$ where $b\in B\setminus \{a\}$, and suppose $B$ is hausdorff and path connected.

Let $C \subseteq B$ be open and simply connected with $b \in C$. Define $\iota ': \pi _1(A,a) \to \pi _1(B,a)$ as $\iota'([\lambda])=[\iota\circ \lambda]$. Clearly $\iota'$ has the morphism property.

Supposing $C\setminus \{b\}$ is simply connected, then prove $\iota'$ is a bijection.

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I am trying to solve this with techniques from Hatcher chapters 0-1.

I have tried to deduce this by considering the implications of a simply connected $C\setminus \{b\}$, namely path connected with trivial $\pi_1$. I have also considered how one might be able to use the hausdorff property of $B$, but it is not clear to me.

In fact it is not even clear that $A$ is path connected with this assumption because one can have a pair of points outside $C\setminus \{b\}$ that are only joined by a path through $b$.

More, if we restrict our attention to the Euclidean plane, then it seems obvious that $C\setminus \{b\}$ cannot be simply connected. So if I am correct, $B$ cannot be a subspace of $R^n$ for $n<3$.

I am not certain how to attack this problem. There seem to be many approaches and I don't see which one is likely to help. Any hint is much appreciated.

Answer 1

If I understand your problem correctly, I think it can be expressed like this:

Let $B$ be a Hausdorff and path-connected topological space with basepoint $a$. Let $b\in B$ such that $b\neq a$ and consider the homomorphism $\iota'\colon\pi_1(B\setminus \{b\}, a) \to \pi_1(B, a)$ induced by inclusion. Show that if $b$ has an open simply-connected neighbourhood $C$ such that $C\setminus \{b\}$ is also simply-connected then $\iota'$ is an isomorphism.

This situation is perfect for van Kampen's theorem. The hint is that $B = B\setminus \{b\} \cup C$. (Note that $B\setminus\{b\}$ is open because $B$ is Hausdorff!)

One part which may be confusing is seeing why $\iota'$ and the isomorphism given by the van Kampen theorem agree. Look carefully at what Hatcher says about free products and how he defines the homomorphism $\Phi$ on pages 42-43.