Disjoint Topological Embeddings

Question

Each canonical injection $\omega_i: X_i \rightarrow \coprod_{i \in I}X_i$ is a topological embedding and an open map.

My attempt:

To proceed, I will identify $X_i$ with $X_i^*= w_i(X_i)$. If $U$ is open in the disjoint union then $w_i^{-1}(U)$ $=$ $U\cap X_i$ is open in $X_i$ for each $i$. Hence it is continuous for each i. Clearly, the function is injective. Hence restricting its codomain to $X_i$ yields the identity map, and since the topologies of the codomain and domain are the same, the map is a homeomorphism onto its image. If $V$ is open in the codomain then for each i$\in I$ $w_i(V)$ $=$ $V$, which is open. Hence the map is an open topological embedding.

Is my attempt correct? (Please answer this)

Answer 1

This is formally incorrect, because formally $X_j\not\subseteq \coprod_{i\in I} X_i$. There's lots of hand-waving when you jump between the disjoint union and its members. You talk about "identifying" $X_j$ with its image $X_j^*$ but that's just pretty much restating that $\omega_j$ is a homeomorphism onto image, right?

Either way, this nuance can be resolved if we just go back to definitions:

$$\coprod_{i\in I}X_i=\bigg\{(i,x)\in I\times\bigcup_{i\in I}X_i\ \bigg|\ x\in X_i\bigg\}$$

The topology on $\coprod_{i\in I}X_i$ is given as follows: $U\subseteq\coprod_{i\in I}X_i$ is open if for any $i\in I$ the set $\{x\in X_i\ |\ (i,x)\in U\}$ is open in $X_i$.

With that the standard embedding $\omega_j:X_j\to\coprod_{i\in I} X_i$ is simply given by $\omega_j(x)=(j,x)$. It's obviously injective. With this function we can also rewrite our topology: $U\subseteq \coprod_{i\in I}X_i$ is open if and only if for any $j\in I$ the set $\omega_j^{-1}(U)$ is open in $X_j$. And so clearly $\omega_j$ is continuous.

On the other hand $\omega_j$ is an open map. This follows from the simple observation that if $U\subseteq X_j$ is open then so is $\{j\}\times U$ in $\coprod X_i$.

Finally an injective, open and continuous map is a homeomorphism onto image. This follows from the following observation: if $f$ is a bijection then $f$ is continuous if and only if $f^{-1}$ is open.

Answer 2

The way freakish presents it, is the cleanest: with a concrete "disjointification" of the $X_i$ so that we can distinguish $X_i$ from its topological copy $\{(i,x): x \in X_i\} \subset \coprod_{i \in I} X_i$.

If we ignore that and treat the $X_i$ as really disjoint from the start and the disjoint sum as a normal union and the $\omega_i$ as "identify-embeddings" we can just say that $\omega_i$ is continuous by definition (the disjoint union topology is the strongest topology that makes all $\omega_i$ continuous), 1-1-ness is obvious, and openness too: as $U \subseteq X_i$ open (and $X_i$ open in $\coprod_{i \in I} X_i$ as $X_i \cap X_j$ is always open (almost almost empty, except when $i=j$ when it's $X_i$, open in $X_i$ so by definition of the union topology we've shown $X_i \subseteq \coprod_{i \in I} X_i$ is open) so $U$ is open in $\coprod_{i \in I} X_i$ and so $\omega_i$ is an embedding.

So your attempt is incorrect in a formal sense, but is sort of correct in an informal one.