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In Appendix A, p. 372 of Rolfen's 'Knots and Links', Rolfsen states the van Kampen theorem, and follows with: "Somewhat more generally, suppose the inclusion homomorphisms $i_{1^\ast}$ and $i_{2^\ast}$ are injective, then one may deduce that $j_{1^\ast}$ and $j_{2^\ast}$ are also injective...", with reference to the commutative diagram attached to this post (this is the usual diagram that accompanies the theorem). I am struggling to figure out why this is the case, I'd be grateful for any advice.

commutative diagram

jgon
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Dario
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    Do you know about amalgamated free products? If so, injectivity follows from the universality property with respect to $G=\pi_1(X_1)*_{\pi_1(X_0)} \pi_1(X_2)$. – Moishe Kohan Feb 10 '20 at 22:05
  • If I have understood it correctly, then by the universality property we have that $g$ is a unique homomorphism. By van Kampen of course, we even have that $g$ is an isomorphism. How does it follow from this that $j_{1^\ast}$ and $j_{2^\ast}$ are injective ? – Dario Feb 11 '20 at 12:01
  • It is pure group theory. One can show that $h_1, h_2$ are injective if $i_{1}, i_{2}$ are injective (but this is not trivial). Thus also $j_{1}, j_{2}$ are injective. See also https://math.stackexchange.com/q/216707 abd Derek Holt's comment. Also see https://homepage.univie.ac.at/bernhard.kroen/bac02fin01.pdf. – Paul Frost Feb 12 '20 at 15:25
  • @PaulFrost Showing the first part of your comment is precisely what I'm struggling with. For it seems that the universality property only comes to use when, having shown that $h_1$ and $h_2$ are injective if $i_{1^\ast}$ and $i_{2^\ast}$ are injective, showing that $j_{1^\ast}$ and $j_{2^\ast}$ are injective as a consequence. – Dario Feb 13 '20 at 15:23
  • So have a look at the linked document. – Paul Frost Feb 13 '20 at 15:28
  • @PaulFrost the only mention of injectivity in the above document i can find is in theorem 4.2. Is this the correct place to look? For I know little to nothing about graph theory. – Dario Feb 14 '20 at 17:59
  • See Corollary 2.7. – Paul Frost Feb 14 '20 at 23:41
  • @PaulFrost the argument seems to go as follows: $h_i$ maps $g \in \pi_1(X_i)$ onto $a \cdot g \in G$, where $a \in \pi_1(X_0)$ and $a \cdot g$ is in normal form (as in definition 2.3). Now if there is some $1 \neq g \in \pi_1(X_i)$ s.t $h_i(g) = a \cdot g = 1 \in G$ then $a =1$ and $g=1$ since $a \cdot g$ is in normal form. Since $g =1$ we have a contradiction and $h_i$ is injective. Is this correct ? – Dario Feb 15 '20 at 13:50

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