$\pi_1$ is a functor from the category of pointed topological spaces to the category of groups.

It's clear that this functor preserves products, since $\pi_1(X\times Y)=\pi_1(X)\times\pi_1(Y)$, and a special case of Van Kampen's theorem states that $\pi_1$ preserves coproducts, since $\pi_1(X\vee Y)=\pi_1(X)*\pi_1(Y)$.

It doesn't seem like too much trouble to show that $\pi_1$ preserves pullbacks, and Van Kampen's theorem helps us with pushouts up to some degree of niceness concerning our spaces.

How many other limits or colimits does $\pi_1$ preserve? I'm not sure I know where to start if I want to talk about equalizers and coequalizers!

If we could show how $\pi_1$ preserves products and equalizers could we conclude that $\pi_1$ preseved all limits (since $Top$ and $Grp$ are both complete and cocomplete)? Could we say the same thing about coproducts?

Are there any obvious counter-examples?

I've had a good look through Hatcher's 'Algebraic Topology' and a quick breeze through JP May's 'Concise Course', but I'm not finding all the answers I'm after.

Any words on this topic would be appreciated :)

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    $\begingroup$ "It doesn't seem like too much trouble to show that $\pi_1$ preserves pullbacks" Are you sure? Also, $\pi_1(X \vee Y) = \pi_1(X) * \pi_1(Y)$ does not always hold (unfortunately), take $X=Y=$ Hawaiian earring. $\endgroup$ Commented Oct 22, 2014 at 9:44
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    $\begingroup$ $\pi_1$ is representable in the homotopy category of pointed spaces, so it preserves all limits in that category. Unfortunately, there aren't so many of those. I would not be so quick to say that it preserves coproducts either; after all, van Kampen's theorem only applies under certain niceness hypotheses. $\endgroup$
    – Zhen Lin
    Commented Oct 22, 2014 at 9:46
  • $\begingroup$ @MartinBrandenburg If we restrict our category of topological spaces to 'nice enough' spaces (as in Hatcher's hypothesis for VKT maybe), then could we talk about coproducts working? $\endgroup$ Commented Oct 22, 2014 at 9:58
  • $\begingroup$ @Zhenlin Is $\pi_1$ only representable in the homotopy category of pointed spaces? you guys are quick! $\endgroup$ Commented Oct 22, 2014 at 9:58
  • $\begingroup$ $\pi_1$ is not representable on $Top_*$ since otherwise it would preserve limits, but it does not preserve pullbacks. $\endgroup$ Commented Oct 22, 2014 at 9:59

3 Answers 3


The doom and gloom here comes from using the wrong notion. Note that limits and colimits of topological spaces are not homotopically invariant (that is, they aren't well-defined operations on homotopy classes of diagrams of spaces), so for the purposes of homotopy theory they aren't always the right thing to do. You should instead be looking at homotopy limits and homotopy colimits, which are in a suitable sense the "derived functors" of limits and colimits; in particular, they are homotopically invariant constructions.

Here the news is much better. The nicest statement comes when we avoid basepoints and instead replace the fundamental group with the fundamental groupoid $\Pi_1(X)$, which is a (higher) functor from spaces to the $2$-category of groupoids, functors, and natural transformations. This is important because $2$-categories also have a notion of homotopy limits and colimits, also called $2$-limits and $2$-colimits. And now I claim that

$\Pi_1$ preserves all homotopy colimits.

This is an abstract and high-powered version of the Seifert-van Kampen theorem (although admittedly it's only useful to the extent that you can actually compute homotopy colimits). Unlike the usual Seifert-van Kampen theorem, it is powerful enough to allow you to compute the fundamental group of $S^1$ by decomposing it as two intervals which intersect in two points; the point is that working with fundamental groupoids gives you the freedom to use more than one basepoint. This argument should be somewhere in Brown's Topology and Groupoids.

Morally this result is true because $\Pi_1$ is the homotopy left adjoint to the "forgetful functor" sending a groupoid $\Pi$ to its classifying space $B\Pi$ (a mild generalization of the construction of Eilenberg-MacLane spaces). The ordinary Seifert-van Kampen theorem can be thought of as giving conditions under which it is possible to compute a homotopy pushout in spaces as an ordinary pushout.

Example. Let $G$ be a group acting on a path-connected space $X$ preserving a basepoint $x$. The ordinary quotient $X/G$, or $X_G$, is a prime example of a colimit which is not a homotopically invariant operation: in general the homotopy type of the quotient is very sensitive to exactly how $G$ acts on $X$, and we cannot replace the action by a homotopy equivalent action. In particular the fundamental group of $X/G$ is not determined from the data of the action of $G$ on $\pi_1(X, x)$.

The homotopically invariant replacement is the homotopy quotient or Borel construction, variously notated $X//G$ or $X_{hG}$, and given explicitly by

$$X \times_G EG$$

where $EG$ is a contractible space on which $G$ acts freely and $\times_G$ means to take the quotient of the product by the diagonal action of $G$. The idea is that $X \times EG$ is a "resolution" of $X$ as a $G$-space suitable for computing the "derived functor" of taking quotients.

The abstract Seifert-van Kampen theorem above now implies that the fundamental group of $X//G$ is the homotopy quotient of $\pi_1(X, x)$ by the action of $G$. Once you work out what homotopy quotients are for groupoids, this turns out to be precisely the semidirect product

$$\pi_1(X, x) \rtimes G.$$

Something more complicated happens if you don't assume that $G$ preserves a basepoint or if you use the more general and homotopically correct definition of "action of $G$," but that's another story.

  • $\begingroup$ (+1) Just wondering, how are you so smart? :) $\endgroup$
    – The Artist
    Commented Oct 26, 2014 at 10:11
  • $\begingroup$ However, the relation between (homotopy) colimits in $\mathbf{Grpd}$ and colimits in $\mathbf{Grp}$ is not so clear. (For instance, think about the groupoid version of the pushout of $\partial \Delta^1 \hookrightarrow \Delta^1$ along itself.) $\endgroup$
    – Zhen Lin
    Commented Oct 26, 2014 at 10:27
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    $\begingroup$ For more on groupoid matters, you could look at . arXiv:1207.6404 Groupoids and Faa di Bruno formulae for Green functions in bialgebras of trees Imma Galvez-Carrillo, Joachim Kock, Andrew Tonks arxiv.org/abs/1207.6404 $\endgroup$ Commented Oct 29, 2014 at 17:05
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    $\begingroup$ A more general question is that people tend it seems to think of spaces, or spaces with base point, and not of spaces with many base points, or of other structured spaces such as filtered spaces. I found these needed to get useful strict higher homotopy groupoids. $\endgroup$ Commented Oct 29, 2014 at 17:10
  • $\begingroup$ @Qiaochu Yuan : Just noticed your remarks on the fundamental group of an orbit space: there is a whole dicussion of the fundamental groupoid of an orbit space in Chapter 11 of Topology and Groupoids. The useful point is that if $G$ acts on $X$ then it also acts on the fundamental groupoid $\pi_1 X$, so one ask: when is the natural morphism from the orbit groupoid $(\pi_1 X)/G \to \pi_1(X/G)$ an isomorphism? Useful circumstances in which it is are given (e,g, $G$ is finite, $X$ is Hausdorff, and has a universal cover). $\endgroup$ Commented Feb 16, 2017 at 14:40

$\pi_1$ preserves arbitrary products (not just finite ones); this is easy to check.

$\pi_1$ does not preserve coproducts in general. See math:SE/320812. Seifert van Kampen's Theorem only applies under certain assumptions.

$\pi_1$ does typically not preserve pushouts. For example $S^1$ is the pushout of two open intervals which have trivial $\pi_1$, but $\pi_1(S^1)$ is not trivial. (A possible replacement for this failure is the long exact sequence of homotopy groups associated to a fibration. Also, we have Seifert van Kampen's Theorem which states that certain "nice" pushouts are preserved.)

$\pi_1$ does typically not preserve pullbacks. For example, $S^1$ is the intersection of two hemispheres $\cong D^2$ in $S^2$, which have trivial $\pi_1$, but $\pi_1(S^1)$ is not trivial.

$\pi_1$ does not preserve monomorphisms (consider $S^1 \to D^2$) and it does not preserve epimorphisms (consider $\mathbb{R} \to S^1$).

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    $\begingroup$ Is $S^1$ the pushout of two intervals in the category of pointed spaces (the context of the question)? I'm not so sure about that. $\endgroup$ Commented Oct 26, 2014 at 10:35
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    $\begingroup$ @Najib: you're right, it's not. The problem is that we have to pick a basepoint in one of the two components of the intersection of the intervals, so the corresponding pushout in pointed spaces ignores the other component. A counterexample for $\pi_1$ not preserving pushouts should involve a badly behaved basepoint. $\endgroup$ Commented Oct 26, 2014 at 10:46
  • $\begingroup$ S^1 should be the pushout in pointed spaces of two inclusions S^0_* --> [0,1]_0 (interval pointed at 0) sending * to 0 and the other point of S^0_* to 1 $\endgroup$
    – Arducode
    Commented Dec 7, 2023 at 18:18

In addition to the other answers, $π_1$ also preserves something for homotopy inverse limits over $ω$ (or any other ordinal sequence).

Let $$X \to \dots \to X_2 \to X_1 \to X_0$$ denote an inverse system of pointed spaces with $X$ being the homotopy limit of the inverse system. Then the natural map $$π_1(X) \longrightarrow \varprojlim_{n < ω} π_1(X_n)$$ is surjective.


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