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Here is the bit of the proof I didn't really understand:

Let $X=X_1\cup X_2$, where $X_1$, $X_2$, $X_1\cap X_2$ and $X$ are path-connected and $X_1, X_2$ are open subsets of $X$. Moreover, assume that the generator set $G_1$ with the relations $R_1$ gives a presentation of $\pi_1(X_1)$ and similarly $G_2, R_2$ gives a presentation of $\pi_1(X_2)$ and $G_1\cap G_2=\varnothing$. Also $i_1:X_1\cap X_2\subset X_1$ and $i_2:X_1\cap X_2\subset X_2$ are the usual imbeddings.

Now here is the statement:

The generator set $G=G_1\cup G_2$ together with the relations $R=R_1\cup R_2\cup R_{12}$ gives a presentation of $\pi_1(X)$, where $R_{12}=\{i_{1\ast}(\alpha)=i_{2\ast}(\alpha):\forall \alpha\in\pi_1(X_1\cap X_2)\}$. Why is $R_{12}$ so strange? What is the idea behind it?

I understood that that new generator set generates $\pi_1(X)$. So it comes next to show that $R$ is the kernel of the homomorphism $\theta:F(G)\to\pi_1(X)$, where $F(G)=F(G_1)\ast F(G_2)$ and $F(G_i)$ are the sets generated by $G_i$. How can I show the inclusion $R\subset \ker(\theta)$?

$R$ consists of words, right? If a word $w\in R_1$ or $R_2$ we are done because $R_1, R_2$ are the kernels of their own homomorphisms $\theta_1$ and $\theta_2$. But what happens when $w\in R_{12}$?

I think I have a wrong picture in my head about relations and presentation.

Xena
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    The idea behind $R_{12}$ is that if you have a loop $\lambda$ in $X_1 \cap X_2$, its class in $\pi_1(X)$ doesn't depend on whether you view it as a loop in $X$ by inclusion via $X_1$ or via $X_2$. So with the inclusions $j_k \colon X_k \to X$, you get $j_{1\ast}\circ i_{1\ast} = j_{2\ast} \circ i_{2\ast}$ as maps $\pi_1(X_1\cap X_2) \to \pi_1(X)$. – Daniel Fischer Aug 23 '13 at 15:18
  • Thank you Daniel. So does it follow that if $w\in R_{12}$ then it is in the kernel just because $w$ is of the form $i_{1\ast}(\alpha)i^{-1}_{2\ast}(\alpha)$ and this is "equal" to the identity? What does it mean that it's equal to the identity? That it's mapped (with what?) to the identity? – Xena Aug 23 '13 at 15:35
  • When you have an instance of $i_{1\ast}(\alpha)i_{2\ast}(\alpha)^{-1}$ in a word, that corresponds to a part $\lambda\lambda^-$ in the loop, where $\lambda$ is a loop in $X_1\cap X_2$ with homotopy class $\alpha$, and $\lambda^-$ is $\lambda$ traversed in the opposite direction. $\lambda\lambda^-$ is already nullhomotopic in $X_1 \cap X_2$, so a fortiori in $X$. Hence $i_{1\ast}(\alpha)i_{2\ast}(\alpha)^{-1}$ can be cancelled from the word. – Daniel Fischer Aug 23 '13 at 15:58
  • Okay, but still it is unclear that $R\subset\ker(\theta)$. Can you help me with that too? Please? – Xena Aug 23 '13 at 16:08
  • perhaps you're missing just this: if you understand relations as words, relation $x=y$ ($x,y$ some words) means, by definition, the word $xy^{-1}$ – user8268 Aug 23 '13 at 20:16
  • @user8268 Alright, a different question: to prove the other inclusion, I should show that if $w\in\ker(\theta)$ then $w\in R$. Does that mean that I should show that $w=1$? – Xena Aug 24 '13 at 12:00

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