5
$\begingroup$

Let $X$ be a topological space, $A$ a contractible subspace of $X$, and $f : X \rightarrow X/A$ the quotient map. I want to say that $f$ always induces an isomorphism between fundamental groups, or at least that it does under certain assumptions (e.g., when $X$ is a CW complex), but I do not see how to show this explicitly except when $X$ is a finite graph and $A$ is a tree in $X$. Is there a way to prove this in general, or at least in a more general case than finite graphs?

$\endgroup$
5
$\begingroup$

There is a result which states:

If the map $i:(A,x_0)\hookrightarrow (X,x_0)$ is a cofibration, and $A$ is contractible then the projection $p:(X,x_0)\to (X/A,\ast)$ is a homotopy equivalence.

In particular, if $(X,A,x_0)$ is a relative CW-complex, then the inclusion $i:(A,x_0)\hookrightarrow (X,x_0)$ is a cofibration. Since homotopy equivalences induce isomorphisms on fundamental groups, the induced map $$p_*:\pi_1(X,x_0)\to\pi_1(X/A,\ast)$$ is an isomorphism in this case.

To show this, let $$H:A\times I\to A,\quad H(a,0)=id_A(a)=a,\quad H(a,1)=x_0$$ be a homotopy giving $A\simeq\ast$. Since $i$ is a cofibration, this can be extended to a homotopy $K:X\times I\to X$ satisfying $$K(x,0)=id_X(x)=x,\quad K(a,1) = H(a,1)=x_0.$$ By the universal property of quotients, the map $K|_{X\times\{1\}}:X\to X$ factors through a map $k:X/A\to X$ such that $K|_{X\times\{1\}}=k\circ p$. Hence, $k\circ p\simeq id_X$.

Next, as $p\circ K|_{A\times I}=\ast$, $K$ factors through a homotopy $\overline{K}:X/A\times I\to X/A$ such that $\overline{K}\circ(p\times id_I)=p\circ K$. It's not hard to verify that $\overline{K}$ gives $id_{X/A}\simeq p\circ k$.

For a reference, everything presented above is in Switzer's Algebraic Topology, chapter 6.

$\endgroup$
7
$\begingroup$

The result isn't true in general.

Take $X=S^1$, $A=S^1 -\{N\}$, $N$ being the north pole. $A$ is contractible and $X/A$ is the Sierpiński space, which is contractible (and thus has a trivial fundamental group), but $X$ has a non-trivial fundamental group.

$\endgroup$

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.