1
$\begingroup$

Consider a path-connected topological space $X$, to which we attach a $1-$cell via the map $\phi:S^0 \to X$, where $S^0 = \{-1,1 \}$. The space we obtain is $$Y = (X \sqcup [-1,1]) / \{-1 \sim \phi(-1) \text{ and } 1 \sim \phi(1) \}. $$

How can we prove that the inclusion $i:X \hookrightarrow{} Y$ induces an injective homomorphism $i_*:\pi_1(X,p) \to \pi_1(Y,p)$ for every $p \in X$?

Intuitively, I see that the attaching a $1-$cell to the space $X$ is similar to $X \lor S^1$ (in the case that we can deformation retract the path between $\phi(1)$ to $\phi(-1)$ in $X$ to a point), so the fundamental group of $Y$ is $\pi_1(X) * \mathbb{Z}$. But what if we cannot deformation retract this path to a point? Also, how do we specifically prove that the inclusion induces an injective homomorphism (we know that it always is a homomorphism); computing the fundamental group of $Y$ using the theorem of Seifert-van Kampen doesn't seem to suffice.

$\endgroup$
1
$\begingroup$

As $X$ is path connected, we can define a continuous map $f\colon Y\to X$ that is the identity on $X$ and maps the glued cell to a path in $X$ between the glue points. Now consider a loop that is in the kernel of $i_*$ and apply $f$ to the corresponding homotopy.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So if $\gamma$ is a loop in $X$ at $p$ with $[i \circ \gamma] = i_*([\gamma])$ being the constant loop in $Y$ at $p$, we want to prove that $\gamma$ is the constant loop in $X$ in $p$. Applying $f$ to the homotopy between $i \circ \gamma$ and the constant loop gives a homotopy between $f\circ i \circ \gamma$ and the constant loop at $f(p)$ (both of them being loops in $X$), but how do we proceed from here? $\endgroup$ – Otp Mar 8 at 10:20
  • $\begingroup$ @Otp We have $f \circ i = id$, thus $f_* \circ i_* = id$. This implies that $i_*$ is injective. $\endgroup$ – Paul Frost Mar 8 at 10:52
  • $\begingroup$ I see now, thank you! $\endgroup$ – Otp Mar 8 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.