Definition:
Suppose $(X_i)_i$ is an indexed family of non-empty topological spaces. There is a canonical injection $\sigma_i: X_i \rightarrow \coprod_{i\in I}X_i$ , given by $\sigma_i(x)=(x,i)$.
The author states that it is a convention to identify $X_i$ with $\sigma_i(X_i)$
Let $X=\coprod_{j\in J}X_j$
The topology is on the disjoint union is defined as
$\tau$ $=$ $\{$ $U\subseteq \ X$ $:$ $U\cap X_j$ (as a subset of X) is open in $X_j$ for each $j\in J$ $\}$
My understanding is as follows:
without identifying the sets, we get that for $U\subseteq X$:
$x\in \sigma_j^{-1}(U) \iff$ $x\in X_j$ and $\sigma_j(x)\in U$ $\iff$ $x\in X_j$ and $(x,j)\in U$ $\iff$ $(x,j)\in \sigma_j(X_j)$ and $(x,j)\in U$ $\iff$ $(x,j)$ $\in \sigma_i(X_j)$ $\cap$ $U$ .
Identifying the sets means that we treat $\sigma_j^{-1}(U)$ as $\sigma_j(X_j) \cap U$ and vice versa.
So, the topology is really defined to be:
$\tau$ $=$ $\{$ $U\subseteq \ X$ $:$ $\sigma_j^{-1}(U)$ is open in $X_j$ for each $j\in J$ $\}$
(Without the identification)
**Is my understanding correct?
May I have advice on how to think about these? Why is the author identifying sets? Why not just write them "normally"?
Best Answer
Yes, you are correct. The topology is the final topology wrt the standard injections $\sigma_i, i \in I$ on the space $\coprod_{i \in I} X_i$ (defined as the standard set disjoint union construction).
It's easy to observe that each separate map $\sigma_{i_0}$ is also open, as $(\sigma_i)^{-1}[\sigma_{i_0}[O]]= O$ for $i=i_0$ and $\emptyset$ if $i \neq i_0$, and when $O \subseteq X_{i_0}$ is open, this set is open for all $i$ for and hence is sum-open in the final topology. So $\sigma_i[X_i]$ is homeomorphic to $X_i$ for all $i$, hence the identification.
And then you can reformulate $O$ being open in $\coprod_{i \in I} X_i$ as
$$\forall i \in I: O \cap \sigma_i[X_i] \text{ open in } \sigma[X_i]$$
because $$O \cap \sigma_i[X_i] = \sigma_i^{-1}[O]$$
So that $\coprod_{i \in I} X_i$ has the so-called coherent topology wrt its subspaces $\sigma_i[X_i]$ (after we've declared $\sigma_i$ to be a homeomorphism, so making the identification).
So this gives two views on the topology, that come down to the same idea in the end.