# the proof of Van Kampen theorem in Hatcher's book

Hatcher defines two moves that can be performed on a factorization of $$[f]$$，The second move is

Regard the term $$[f_i]\in\pi_1(A_\alpha)$$ as lying in the group $$\pi_1(A_\beta)$$ rather than $$\pi_1(A_\alpha)$$ if $$f_i$$ is a loop in $$A_\alpha\cap A_\beta$$.

In the book,the author mentions that the move does not change the image of this element in the quotient group $$Q=\ast_\alpha\, \pi_1(A_\alpha)/N$$,by the definition of $$N$$,how to conclude the above conlusion,can anyone show me any examples to interpret the statement?

If we have $$X=A_1\cup A_2\cup A_3$$,$$[f_1]\in \pi_1(A_1\cap A_3),[f_2]\in \pi_1(A_2)$$,does it mean that $$[f_1][f_2]$$ and $$[f_2][f_1]$$ are equivalent?

$$N$$ is the normal subgroup of $$*_\alpha \pi_1(A_\alpha)$$ generated by all words of the form $$i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{-1}$$ with $$\omega \in \pi_1(A_\alpha \cap A_\beta)$$.

Hatcher does not formulate the second move very precisely. He "really" means that if $$g_j$$ is a loop in $$A_\alpha \cap A_\beta$$, i.e. $$[g_j] \in \pi_1(A_\alpha \cap A_\beta)$$, then we are allowed to replace $$[f_j] = i_{\alpha\beta}([g_j]) \in \pi_1(A_\alpha)$$ by $$[f'_j] = i_{\beta\alpha}([g_j]) \in \pi_1(A_\beta)$$.

To understand this, note that by definition the word $$[f_j][f'_j]^{-1}$$ is contained in $$N$$. Similarly, since $$[g_j]^{-1} \in \pi_1(A_\alpha \cap A_\beta)$$, we see that also $$[f_j]^{-1}[f'_j] = i_{\alpha\beta}([g_j])^{-1}i_{\beta\alpha}([g_j]) =$$ $$i_{\alpha\beta}([g_j]^{-1})i_{\beta\alpha}([g_j]^{-1})^{-1} \in N$$. This implies that if $$w$$ is an arbitrary word containing $$[f_j]^\epsilon$$, $$\epsilon = \pm 1$$, and the word $$w'$$ is obtained by replacing $$[f_j]^\epsilon$$ with $$[f'_j]^\epsilon$$, then $$w,w'$$ have the same equivalence class in $$Q$$. To see this, note that we can write $$w = u [f_j]^\epsilon v$$ and $$w' = u [f'_j]^\epsilon v$$ with suitable words $$u, v$$ in $$*_\alpha \pi_1(A_\alpha)$$. Then $$w(w')^{-1} = u [f'_j]^\epsilon v v^{-1} [f'_j]^{-\epsilon}u^{-1} = u [f'_j]^\epsilon [f'_j]^{-\epsilon}u^{-1} \in N$$ because $$[f'_j]^\epsilon [f'_j]^{-\epsilon} \in N$$ and $$N$$ is a normal subgroup of $$*_\alpha \pi_1(A_\alpha)$$.

In your example you have $$[g]\in \pi_1(A_1\cap A_3)$$, $$[f_1] = i_{13}([g]) \in \pi_1(A_1)$$, $$[f_2]\in \pi_1(A_2)$$. With $$[f_3] = i_{31}([g]) \in \pi_1(A_3)$$ you see that $$[f_1][f_2]$$ and $$[f_3][f_2]$$ are equivalent, but it is not true that$$[f_1][f_2]$$ and $$[f_2][f_1]$$ are equivalent.

Edited:

Due to the discussion in the comments let me "go back to the roots".

A map $$f : X \to Y$$ is determined by its domain $$X$$, its range $$Y$$ and the rule of assignment $$X \ni x \mapsto f(x) \in Y$$. The range is an essential constituent of $$f$$. If you have a space $$Y' \supset Y$$, you may regard $$f$$ as map into $$Y'$$, but the precise interpretation is that you replace $$f$$ by the map $$i \circ f : X \to Y'$$ (where $$i : Y \to Y'$$ denotes the inclusion map). The maps $$f$$ and $$i \circ f$$ are not the same because the have different ranges (although you have $$f(x) = (i \circ f)(x)$$ for all $$x \in X$$). Conversely, if you have a map $$f' : X \to Y'$$ such that $$f(X) \subset Y$$, you may regard $$f$$ as map into $$Y$$, but the precise interpretation is that there exists a unique map $$f : X \to Y$$ such that $$f' = i \circ f$$.

Precision is no nitpicking. This is of particular importance if you consider homotopy classes of maps. The range determines how much space is available for deformations. Two non-homotopic maps $$f, g : X \to Y$$ may become homotopic in a bigger $$Y' \supset Y$$ which means that $$i \circ f, i \circ g$$ are homotopic. As an example take $$id : S^1 \to S^1$$ which is not null-homotopic, but becomes null-homotopic in $$\mathbb R^2$$.

Hatcher does not address this point, probably it seems self-evident to him. When he says that $$g_{k-1} \circ f \circ g_k$$ is a loop in $$A_k$$, he means that the map $$g_{k-1} \circ f \circ g_k : I \to X$$ has the property $$(g_{k-1} \circ f \circ g_k)(I) \subset A_k$$ and thus can be written as $$g_{k-1} \circ f \circ g_k = \iota_k \circ \phi_k$$ with $$\phi_k : I \to A_k$$, where $$\iota_k : A_k \to X$$ denotes the inclusion map.

Then $$[\phi_k] \in \pi_1(A_k)$$ and you get $$i_k([\phi_k]) = (\iota_k)_*([\phi_k]) = [\iota_k \circ \phi_k] = [g_{k-1} \circ f \circ g_k] \in \pi_1(X)$$.

• But $i_{\alpha\beta}$ and $i_{\beta\alpha}$ are inclusion maps, $i_{\alpha\beta}([g_j]) =[g_j],i_{\beta\alpha}([g_j])=[g_j]$. Commented Sep 27, 2019 at 23:52
• No. $[g_j]$ is an element of $\pi_1(A_\alpha \cap A_\beta)$, but $i_{\alpha\beta}([g_j]) \in \pi_1(A_\alpha)$ which is different group. I denoted the lattter by $[f_j]$. By an abuse of notation you can write $[g_j]$ instead of $[f_j]$. This is Hatcher's imprecision. But $\pi_1(A_\alpha \cap A_\beta)$ is definitely not a genuine subgroup of $\pi_1(A_\alpha)$. Commented Sep 28, 2019 at 0:10
• $\pi_1(A_\alpha \cap A_\beta)$ is not a subgroup of $\pi_1(A_\alpha)$? But according to the definition of fundamental group,any loop $f$ in $A_\alpha \cap A_\beta$ is also a loop in $A_\alpha$,so any homotopy class of $f$ in $\pi_1(A_\alpha \cap A_\beta)$ is also in $\pi_1(A_\alpha)$?Would you mind showing me a counterexample? Commented Sep 28, 2019 at 5:35
• You can of course regard any loop in $A_\alpha \cap A_\beta$ as a loop in $A_\alpha$. The precise meaning is to use the inclusion-induced homomorphism $i_{\alpha\beta} : \pi_(A_\alpha \cap A_\beta) \to \pi_1(A_\alpha)$. But the elements $[g_j]$ and $[f_j] = i_{\alpha\beta}([g_j])$ are not the same and deserve a distinct notation. Note that in general $i_{\alpha\beta}$ is not an injection so that you cannot reconstruct $[g_j]$ from $[f_j]$. Take for example $A_1$ = open disk centered at $0\in \mathbb R^2$ and $A_2$ = $\mathbb R^2 \setminus \{0\}$:$\pi_1(A_1 \cap A_2) =\mathbb Z,\pi_1(A_1)=0$. Commented Sep 28, 2019 at 7:56
• Thank you for your patience! Commented Sep 28, 2019 at 10:36