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I am having difficulty understanding the first homomorphism in the van Kampen Theorem (see Hatcher pp 42-43). I will illustrate my confusion with the simple example of an annulus ($X$), which will be covered by two subsets ($A$ and $B$), each of which include the central opening and constructed so that their intersection is path connected, and so that each contains the base point.

The fundamental group of both $A$ and $B$ is $\mathbb{Z}$. The free product of the fundamental groups of $A$ and $B$ is the argument of the homomorphism, and this free product is $\mathbb{Z}\star\mathbb{Z}$.

Now, my question: How does the homomorphism map $\mathbb{Z}\star\mathbb{Z}$ to $\mathbb{Z}$, the fundamental group of the annulus?

If I understand the definitions correctly, a word $g_1g_2$ in the argument will be mapped to $g_1g_2$. But the range of the homomorphism is $Z$, the fundamental group of $X$, and $g_1g_2$ is not in $Z$.

(P.S. I see, of course, that words like $g_1g_2$ can be reduced with the knowledge that there is only one hole in $X$, and so there never were two loops after all. But this presupposes an understanding of the structure of $X$. And, if one already understands $X$, there is no reason to resort to the van Kampen Theorem in the first place.)

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    $\begingroup$ Could you clarify? What exactly are $A$ and $B$? Cocentric annuli? What do you mean by mapping the word $g_1 g_2$ to $g_1g_2$, if you then claim that $g_1 g_2$ is not in $\mathbb{Z}$? Also, note that van-Kampen involves the intersection of $A$ and $B$, which seems to be absent from your question. $\endgroup$ Aug 15 '20 at 19:32
  • $\begingroup$ Think of A and B as being almost the same as the annulus, but missing a sliver on the left or the right. The map of g1g2 that I refer to is the homomorphism that Hatcher refers to in the first sentence of his statement of the theorem. I note that the intersections are path-connected, and both cover the base point and the opening in the center. Otherwise, the intersections are not used in the first sentence. $\endgroup$
    – PossumP
    Aug 15 '20 at 20:09
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Here is a step-by-step execution of your example:

Seifert-van Kampen (as in Hatcher, adapted). If $X$ is the union of path-connected open sets $A$ and $B$ each containing the basepoint $x_0 \in X$ and if $A\cap B$ is path-connected, then the homomorphism $\Phi : \pi_1(A)∗\pi_1(B)\to\pi_1(X)$ is surjective. The kernel of $\Phi$ is the normal subgroup $N$ generated by all elements of the form $i_{AB}(\omega)i_{BA}(\omega)^{−1}$ for $\omega\in \pi_1(A\cap B)$, and hence $\Phi$ induces an isomorphism $\pi_1(X) \cong \pi_1(A)∗\pi_1(B)/N$. Here, $i_{AB}$ is the homomorphism induced by the inclusion of $A\cap B$ into $A$, and similar for $i_{BA}$.

(Note that the second sentence originally has the condition that the triple intersections are path-connected, which is trivial here.)

In general, this is good news because for example, when your intersection is simply connected, your $i_{AB}$ and $i_{BA}$ must be trivial maps, so $N$ must be trivial. What you have to know is not $\pi_1(X)$ here, but rather $\pi_1(A\cap B)$. In your example, these are isomorphic, whence it is actually not too exciting for that special case.

Now to the particulars: What we have to know is $\pi_1(A)\cong \mathbb{Z}$, $\pi_1(B)\cong \mathbb{Z}$, $\pi_1(A\cap B)\cong \mathbb{Z}$. Let $a$ be a generator of $\pi_1(A)$, $b$ a generator of $\pi_1(B)$. These can be thought as fixed loops, as Michael described, say counterclockwise. Then the elements of $\pi_1(A)*\pi_1(B)$ look like, for instance, $\emptyset$, $aaabbba$, $abbaba$, $a^-b^-aab$, etc. Here with $a^-$ I denote $(-1)a$ (inverse), and if you feel more comfortable, you may write $a^3$ instead of $aaa$, etc.

Now the question is: What does $i_{AB}:\pi_1(A\cap B)\to\pi_1(A)$ do? All loops in $A$ are homotopic to some loop in $A\cap B$ relative endpoint (again, sort of boring because $A$ and $A\cap B$ look "too alike" for it to be interesting), so this map is an isomorphism. Take $\omega$ to be the preimage of $a$. We can easily see that $i_BA(\omega)=b$ (here we "use the knowledge" from $B$ and not from $X$!) So, the same applies for $i_{BA}$ and $\pi_1(B)$, except now we are taking the inverse of $i_{BA}(\omega)$ into account. when calculating $N$. $N$ is generated by $ab^-$, so we will "shorten" by $ab^-=0$, which you can basically think of as not differentiating between $a$ and $b$ (since $a=ab^-b\simeq b$). Then evidently, the free group modulo this identification is $\cong \mathbb{Z}$, given by word length after the identifications.

Summary: You indeed need to know the fundamental group structures of your two subsets that cover your space, and of their intersection. In this particular example, it is tempting to think that we are using the whole space's knowledge where we actually always use that of either one part, or the intersection. But all those groups/spaces were really similar here, so Seifert - van Kampen would be a real overkill. :)

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  • $\begingroup$ Got it. Thanks. $\endgroup$
    – PossumP
    Aug 17 '20 at 23:58
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Let $g_1, g_2$ be the generators of $\pi_1(A)$, $\pi_1(B)$ given by counterclockwise loops around the central hole, and let $h$ be the generator of $\pi_1(X)$ given by a counterclockwise loop around the central hole. Then the map $\pi_1(A)\ast\pi_1(B) \to \pi_1(X)$ induced by the inclusions $A \hookrightarrow X$ and $B \hookrightarrow X$ is given by $g_1 \mapsto h$, $g_2 \mapsto h$. Therefore, the word $g_1g_2$ is mapped to $h^2$.

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  • $\begingroup$ I see this, but it requires prior knowledge of the FG and nature of X. So, what is van Kampen telling me that I didn’t know before? $\endgroup$
    – PossumP
    Aug 15 '20 at 20:04
  • $\begingroup$ Well this example you've chosen isn't particularly insightful. The theorem is usually employed to calculate the fundamental group of $X$ (which you don't already know) in terms of $\pi_1(A)$ and $\pi_1(B)$. $\endgroup$ Aug 15 '20 at 20:16
  • $\begingroup$ Right, but even this simple example requires that I already know the FG of X. In more complicated cases, it would be impossible. (I’m not trying to be argumentative, I’m just trying to really understand what the first sentence of the theorem is telling us.) $\endgroup$
    – PossumP
    Aug 15 '20 at 20:23
  • $\begingroup$ I disagree that you need to know what $\pi_1(X)$ is. The first sentence of Theorem 1.20 tells you that $\pi_1(A)\ast\pi_1(B)$ surjects onto $\pi_1(X)$. By the first isomorphism theorem, you only need to find the kernel to determine $\pi_1(X)$, and this is described by the second sentence of Theorem 1.20. In this case, it tells you that the kernel is the normal subgroup generated by $g_1g_2^{-1}$. Hence, $\pi_1(X)$ is isomorphic to the group $\langle g_1, g_2 \mid g_1 = g_2\rangle \cong \langle h\rangle \cong \mathbb{Z}$. $\endgroup$ Aug 15 '20 at 20:41
  • $\begingroup$ Maybe I am trying to get too much out of the first sentence, before moving onto the rest of the Theorem. $\endgroup$
    – PossumP
    Aug 16 '20 at 23:03

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