Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I prove the following:

If $X\supseteq K$ is contractible, then the quotient $X/K$ is homotopy equivalent to $X$?

Since $K$ is contractible, we have a homotopy $H:id_K\!\simeq\!c_{k_0}$ between the identity and the constant map $K\rightarrow\{k_0\}\!\subseteq\!K$. We are trying to find $f:X\rightarrow X/K$ and $g:X/K\rightarrow X$, such that $f\circ g\simeq id_{X/K}$ and $g\circ f\simeq id_X$.

Most probably, $f$ must be the quotient projection, but what about $g$? I thought about defining $g$ as $x\!\in\!X\setminus K\mapsto x$ and $k\!\in\!K\mapsto k_0$, but is this continuous? Probably not. What else then?

Theorem: The fundamental group of a finite graph $X$ is free.

Proof: if $T$ is the maximal (spanning) tree of $X$, and $E$ the set of edges ($1$-cells) not in $T$, then $T$ is contractible and $X$ is homotopy equivalent to $X/T$, which is (homeomorphic to) a bouquet of circles $\bigvee_{i=1}^{|E|}\mathbb{S}^1$. Therefore $\pi_1(X)\cong F_E$, the free group on the set $E$. $\blacksquare$

enter image description here

Are infinite trees also contractible?

share|cite|improve this question
up vote 10 down vote accepted

The statement "if $K \subset X$ and $K$ is contractible, then $X/K$ is homotopy equivalent to $X$" does not hold in general. Specifically, you need to add the assumption that the pair $(X,K)$ has the homotopy extension property (see Hatcher's algebraic topology book, proposition 0.17). In particular, it works if $X$ is a CW complex and $K$ is a subcomplex.

For a simple counterexample, let $X$ be a circle, and let $K$ be the complement of a single point in $X$. Then $K$ is contractible, and $X/K$ is a non-Hausdorff space that is not homotopy equivalent to a circle.

Your proof that the fundamental group of a finite graph is free is correct, assuming that the graph is connected.

It is also true that any infinite tree is contractible, for the same reason that a finite tree is contractible (you can contract to any point using geodesic paths in the tree). Thus the fundamental group of an infinite graph is also free.

share|cite|improve this answer

You can also generalise the result on graphs to say that the fundamental groupoid of a graph on the set of vertices is a free groupoid; the notion of groupoid is quite natural in $1$-dimensional homotopy theory, and is also a natural extension of group theory. Their use often avoids the use of choice of base point and of tree.

For the notion of free groupoid, see Chapter 4 of the downloadable

Higgins, P.J. Notes on categories and groupoids, Mathematical Studies, Volume 32. Van Nostrand Reinhold Co. London (1971); Reprints in Theory and Applications of Categories, No. 7 (2005) pp 1--195.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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