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I'm studying Algebraic Topology, and I'm thinking about the meaning of $\pi_0(X, A, x_0)$. In Hatcher's exercise, $\pi_0(X, A, x_0):= \pi_0(X, x_0) / \pi_0 (A, x_0)$. But, I don't know how to quotient these set. $\pi_0 (A, x_0)$ is path component of $A$ and $\pi_0(X, x_0)$ is path component of $X$. So I think that $X_\beta = \cup A_{\beta, \alpha}$ that $A_{\beta, \alpha}$ is path component elements. Then, what is $\pi_0(X, A, x_0)$? Is this the member of $\pi_0(X)$ that does not have intersection with $A$?

probafds123
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You are looking at a mistake in an older version of the book. Hatcher is a tiny bit imprecise with notation and I think intended for this to be understood from context. Consider the inclusion $i:A\to X$, which induces a function $i_{\#}:\pi_0(A,x_0)\to \pi_0(X,x_0)$. To extend the long exact sequence you need to define $\pi_0(X,A,x_0)$ as the quotient set $\pi_0(X,x_0)/i_{\#}(\pi_0(A,x_0))$.

The function $\pi_0(X,x_0)\to\pi_0(X,x_0)/i_{\#}(\pi_0(A,x_0))$ is surjective by construction and the sequence is also exact at $\pi_0(X,x_0)$.

This is correct in an updated version of the book.

Jeremy Brazas
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  • Apart from your use of $i_#$ rather than $i_$, what you write is exactly what Hatcher writes (the OP has omitted $i_$). In the context of the exercise, it can only mean $\pi_0(X, x_0)/~$ where $~$ is the equivalence relation which identifies all the points in $i_*(\pi_0((A, x_0))$, which, I imagine, is what you have in mind too. – Rob Arthan Jun 05 '21 at 14:53
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    @RobArthan In the version I have, in Exercise 7 on p. 358, Hatcher does write $\pi_0(X,x_0)/\pi_0(A,x_0)$, which is not correct. It looks like this is corrected in a newer version of the book. – Jeremy Brazas Jun 05 '21 at 15:06
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    I see. It has been corrected in the online version: https://pi.math.cornell.edu/~hatcher/AT/AT.pdf. – Rob Arthan Jun 05 '21 at 15:08
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A pointed set $(P,p_0)$ is a pair consisting of a set $P$ and a point $p_0 \in P$ (which is called the basepoint).

$\pi_0(Y, y_0)$ is such a pointed set which consists of the set of path components of $Y$ with basepoint being the path component $C(y_0;Y)$ of $y_0$ in $Y$.

Each map $f : (Y,y_0) \to (Z,z_0)$ induces a basepoint-preserving function $f_* : \pi_0(Y, y_0) \to \pi_0(Z, z_0)$ by assigning to the path component of $y \in Y$ the path component of $f(y) \in Z$.

The inclusion $i : (A,x_0) \hookrightarrow (X,x_0)$ induces $i_* : \pi_0(A, x_0) \to \pi_0(X, x_0)$, but in general $i_*$ is not injective. As an example consider $(A,x_0) = (\mathbb Z, 0)$ and $(X,x_0) = (\mathbb R, 0)$. Thus it is a severe abuse of notation to define $\pi_0(X, A, x_0):= \pi_0(X, x_0) / \pi_0 (A, x_0)$. The correct variant is $$\pi_0(X, A, x_0):= \pi_0(X, x_0) /i_*( \pi_0 (A, x_0)) $$ which is again a pointed set. This is obtained from $\pi_0(X, x_0)$ by collapsing the subset $i_*( \pi_0 (A, x_0))$ to a single point which becomes the basepoint of $\pi_0(X, A, x_0)$. Therefore $\pi_0(X, A, x_0)$ can be identified with the pointed set $$\left((\pi_0(X, x_0) \setminus i_*( \pi_0 (A, x_0)) \cup \{*\},*\right)$$

where $*$ stands for the point obtained by collapsing $i_*( \pi_0 (A, x_0))$. But clearly $$\pi_0(X, x_0) \setminus i_*( \pi_0 (A, x_0)) = \{ C \in \pi_0(X, x_0) \mid C \cap A = \emptyset \} .$$ Moreover, we may identify $*$ with $C(x_0;X)$ which is certainly not in $\pi_0(X, x_0) \setminus i_*( \pi_0 (A, x_0))$. In other words, we can write $$\pi_0(X, A, x_0) = \{ C \in \pi_0(X, x_0) \mid C \cap A = \emptyset \text{ or } x_0 \in C \} .$$

Paul Frost
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