Does the inclusion of a 1-dimensional space into another induce isomorphism on fundamental groups if the spaces have fundamental group $\mathbb{Z}$?

Question

Let $A \subset B$ be 1-dimensional CW-complexes that both have fundamental group $\mathbb{Z}$. Does the inclusion $A \hookrightarrow B$ induce an isomorphism on fundamental groups?

Corollary 3.3 from https://www.sciencedirect.com/science/article/pii/S0166864105002907 says that the inclusion induces an injection on fundamental groups, but I have not found a reference that says that we have a surjection in this special case.

Answer 1

If for example, $B$ is a finite $1$-dimensional CW-complex, and $A$ is a sub-complex, then you know that you can get $B$ from $A$ by attaching a finite number of $0$-cells and $1$-cells. You also know that you can attach them one by one.

Say you attach every $0$-cell first.

Then it is a induction on the number of $1$-cells : for $n=0$, $A=B$ so their group is isomorphic. Suppose you are at some step of the induction, and that the space you have is $A^\prime$, with the inclusion $A\subset A^\prime$ inducing an isomorphism of the $\pi_1$s. Then, if you attach one $1$-cell, there are two cases : either the map $\varphi:S^0=\{-1, 1\}\to A^\prime$, equivalent to the data of $x,\ y\in A^\prime$ is such that $x=y$, either $x\neq y$. If $x=y$, then we add a copy of $S^1$ to $A^\prime$. If $x=y$ is in $A^\prime$, Van Kampen says that there is a copy of $\mathbb Z * \mathbb Z$ in the new $\pi_1$ which is not possible by your injectivity argument. So $x=y$ are not in $A^\prime$ and we are adding a circle elsewhere. This doen't affect the $\pi_1$ based on a point of $A^\prime$. Now, if $x\neq y$, this can happen that we are linking $A^\prime$ to a circle we added elsewhere at some previous step of the induction. But it is easy to see that we again have a $\pi_1$ which is strictly bigger than $\mathbb Z$, impossible by the injectivity argument. So, we are just adding a line which can be contracted, so the inclusion still induces an isomorphism on the $\pi_1$s. We are done.

Answer 2

After some thinking, I believe I have proven the more general statement:

Let $i : A \hookrightarrow B$ be an embedding of spaces, where $A$ is compact, connected, locally connected metric, and $B$ is compact metric, and with $B \backslash A$ one-dimensional. If $\pi_1(A) = \pi(B) = \mathbb{Z}$, then $i$ induces an isomorphism on fundamental groups.

Proof:

The proof uses Lemma 3.1 from https://www.sciencedirect.com/science/article/pii/S0166864105002907 which says that $B$ retracts to $A$. This means that the composition $A \hookrightarrow B \to A$ is the identity on $A$. The induced composition on fundamental groups $\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}$ is then the identity on $\mathbb{Z}$, so the first map must be an isomorphism.