# Three questions about Wikipedia's definition of Van Kampen's theorem for fundamental groups

I'm studying Algebraic Topology off of Hatcher and (unfortunately as usual) I find his definition and explanation of Van Kampen's theorem to be carelessly written and hard to follow. I happen to know a bit of category theory, so this Wikipedia definition of it seems much easier in principal to understand.

Unfortunately I still can't understand every aspect of this definition (although I still find it much easier to understand the "big picture" of than with Hatcher). The objects and arrows $$\pi_1(U_1 \cap U_2)\xrightarrow {i_\star} \pi_1(U_\star)$$ and $$\pi_1(U_\star) \xrightarrow{j_\star} \pi_1(X)$$ make sense to me, in that the structure of these group homomorphisms are induced by inclusions in the topological spaces.

I have three questions, though:

(1)

In this diagram, is the pushout given as $$\pi_1(X)$$?

(2)

Are the dotted arrows all indicating unique morphisms (that is unique homomorphisms) between groups?

(3)

What is the $$\pi_1(U_1) *_{\pi_1 (U_1 \cap U_2)} \pi_1 (U_2)$$ object in the middle of the pushout diagram meant to indicate? I'm reading Hatchers textbook and he generally uses $$*_\alpha$$ to indicate taking the free product across some index $$\alpha$$. Here however the index $$*_{\pi_1 (U_1 \cap U_2)}$$ is strange to read and/or understand. Finally, you see some kind of concatenation with the $$\pi_1(U_\star)$$'s on the outside and the $$(*_\star)$$ on the inside. I have absolutely no idea what this concatenation (and therefore object with its arrows) represents besides of course the sort of abstract cone point that would invoke the universal property for the pullback.

• Well, that wikipedia argument has its own particular sloppiness. The pushout is not written as an amalgamated free product unless the homomorphisms $i_*$ and $j_*$ are injective. On the other hand Van Kampen's Theorem holds whether or not $i_*$ and $j_*$ are injective. So Hatcher's statement has the virtue of being mathematically correct. Commented Aug 25, 2023 at 0:37

(1) the pushout is $$\pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2)$$ and $$k$$ is the unique morphism provided by the universal property. However, in this case $$k$$ is an isomorphism making $$\pi_1(X)$$ a pushout as well.
(2) the dashed arrows show which morphisms aren't automatically there. The $$i_j\colon\pi_1(U_1\cap U_2)\to\pi_1(U_j)$$ are clearly there, as they are induced by the inclusions $$U_1\cap U_2\to U_j$$, but the rest need some work. First, we know that a free product $$G*H$$ is a coproduct of $$G, H$$ in the category of groups, so it comes with natural morphisms $$G\to G*H, H\to G*H$$. Furthermore, quotienting $$G*H$$ by a normal subgroup $$N$$ gives the natural morphisms $$G\to G*H\to (G*H)/N, H\to(G*H)/N$$. This is how we get the two diagonal dashed arrows (explained further in part (3)), while $$k$$ follows from showing that we do indeed have a pushout diagram and that the solid part of it commutes ($$j_1\circ i_1= j_2\circ i_2$$).
(3) now what actually is $$\pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2)$$ and how is it related to $$\pi_1(U_1)*\pi_1(U_2)$$? Denote by $$\lambda_i$$ the natural morphism $$\pi_1(U_i)\to\pi_1(U_1)*\pi_1(U_2)$$. The question goes: is it true that $$\lambda_1\circ i_1=\lambda_2\circ i_2$$? The answer in general is "nope", but this is a problem if we want a pushout diagram. Hence we want to force the composites to be the same, and for this reason we need to quotient something out. Let $$[\gamma]\in\pi_1(U_1\cap U_2)$$, so that $$(\lambda_1\circ i_1)([\gamma])\left((\lambda_2\circ i_2)([\gamma])\right)^{-1}=i_1([\gamma])i_2([\gamma])^{-1}$$ is in the free product. Generate a normal subgroup $$N$$ by all such elements and quotient that out to get the natural projection $$p\colon\pi_1(U_1)*\pi_1(U_2)\to(\pi_1(U_1)*\pi_1(U_2))/N$$, satisfying $$p\circ\lambda_1\circ i_1 = p\circ\lambda_2\circ i_2.$$ As you may have guessed by now, $$\pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2):=(\pi_1(U_1)*\pi_1(U_2))/N$$, and the composites $$p\circ\lambda_i$$ are the diagonal dashed arrows. One then verifies that this is indeed a pushout and we are done.