Consider $S^1 \hookrightarrow M_f$$i: S^1 \hookrightarrow M_f$ where $f:S^1 \rightarrow S^1$ is squaring and $M_f$ denotes the mapping cylinder. Then the inclusion induces a map $\pi_1(S^1) \rightarrow \pi_1(M_f)$ which has image $2 \mathbb{Z}$. Attaching a 2 dimensional disk to a graph hasAdding additional segments and attaching disks is the effect of quotienting out bysame as adding generators and relations to the normal subgroup generated bypresentation $\langle x,y | y=2x \rangle$ and the map ofgoal is to end up with $y$ generating the boundaryentire group with $|y|=\infty$. For the inclusion to remain an isomorphism, thisThis means that any 2-disk we attach must be through a null homotopic map, since we are after a quotient ofhave $\mathbb{Z}$ isomorphic to$x=ny=2nx$ which implies $\mathbb{Z}$$x$ has finite order which in turn implies $y$ has finite order. Obviously attaching a disk like this does nothingThis means that we cannot attach segments and disks to solvemake the problem, so this is a counter exampleinclusion induce an isomorphism.
