Given a collection of topological spaces $$ \mathfrak X:=\{X_\sigma:\sigma\in\Sigma\} $$ we let consider the natural inclusion $$ \iota_\sigma: X_\sigma\ni x_\sigma\longrightarrow(x_\sigma,\sigma)\in\coprod_{\sigma\in\Sigma}X_\sigma $$ Well I would like to understand if a function $f$ form $\coprod_{\sigma\in\Sigma}X_\sigma$ to a space $Y$ is continuous if and only if $f\circ\iota_\sigma$ is continuous.
So on this purpouse I first observed that the map $$ e_\sigma:X_\sigma\ni x_\sigma\longrightarrow(x_\sigma,\sigma)\in X_\sigma\times\{\sigma\} $$ is an embedding so that I observed that if $$ i_\sigma:X_\sigma\times\{\sigma\}\ni (x_\sigma,\sigma)\longrightarrow(x_\sigma,\sigma)\in\coprod_{\sigma\in\Sigma}X_\sigma $$ is the inclusion then the identity $$ \iota_\sigma=i_\sigma\circ e_\sigma $$ holds and thus I concluded that $\iota_\sigma$ is a continuous function. So if $f$ is continuous then clearly even $f\circ\iota_\sigma$ is continuous. Conversely if $f\circ \iota_\sigma$ is continuous then I observed that $$ f\circ i_\sigma=\big(f\circ (i_\sigma\circ e_\sigma)\big)\circ e^{-1}_\sigma=(f\circ\iota_\sigma)\circ e^{-1}_\sigma $$ for any $\sigma\in \Sigma$ so that by the universal continuous property of final topolgy I concluded that $f$ is continuous.
So I would like to know if my proof attempt is correct and then I would like to know if the coproduct topology is just the final topology induced by the collection $$ I:=\{\iota_\sigma:\sigma\in \Sigma\} $$ I point out that I think that the last statement is false because e.g. if $X_{\sigma_1}$ and $X_{\sigma_2}$ are not disjoint then $\iota^{-1}_{\sigma_2}[A_{\sigma_1}\times\{\sigma_1\}]$ cannot be open in $X_{\sigma_2}$ when it is open in $X_{\sigma_1}$: however I was not able to find an appropriate counterexample. So could someone help me, please?
Yes, your proof is correct. It's good you have seperated the inclusions (and also haven't forgotten the $-\times\{\sigma\}$), but you don't need to when you are more stable.
The coproduct topology (or also called disjoint union topology) is indeed the final topology induced by all inclusions $\iota_\sigma$, which are then all continuous by definition. The forward direction is then simple, as all $f\circ\iota_\sigma$ are simply compositions of continuous maps. For the backwards direction, you could also go another way without using the universal property, but it is related of course. It would look something like this:
Lemma: For topological spaces $X$, $Y$ and $Z$ as well as maps $i\colon X\rightarrow Y$ and $f\colon Y\rightarrow Z$, we have: If $f\circ i$ is continuous and $i$ is open, then $f\vert_{i(X)}$ is continuous.
Let $U\subseteq Z$ be open, then $i^{-1}(f\vert_{i(X)}^{-1}(U))=(f\vert_{i(X)}\circ i)^{-1}(U)\subseteq X$ is open since $f\circ i=f\vert_{i(X)}\circ i$ is continous, therefore $f\vert_{i(X)}^{-1}(U)=f\vert_{i(X)}^{-1}(U)\cap i(X)=i(i^{-1}(f\vert_{i(X)}^{-1}(U))\subseteq Y$ is open since $i$ is open. Therefore $f\vert_{i(X)}$ is continuous. $\square$
Lemma: The inclusion $\iota_\sigma\colon X_\sigma\hookrightarrow\bigsqcup_{\sigma\in\Sigma}X_\sigma$ is open.
Proof: Let $U\subseteq X_\sigma$ be open, then $\iota_\sigma^{-1}(\iota_\sigma(U))=U$ since $\iota_\sigma$ is injective. Therefore we have that $\iota_\sigma(U)\subset\bigsqcup_{\sigma\in\Sigma}X_\sigma$ is open due to the definition of the coproduct topology as the finest (with the most open sets possible), so that all inclusions $\iota_\sigma$ are continuous. $\square$
Therefore, we can conclude that if all $f\circ\iota_\sigma$ are continuous, then all limitations $f\vert_{\iota_\sigma(X_\sigma)}$ are and therefore $f$ is due to the pasting lemma.