Simple! (just add a few edges to connect!) while this may appear to be an easy fix and can approximate a solution in some cases- it is not worthy of a complete proof (as again,it is not comprehensive..

Following perhaps (example 1.11 in the book ((theory of computation , by michael sipser)) ) you can join two languages , but it is a special case as either subset langauge starts with a different letter - consider they were the same like the following two languages from {a,b,c} alphabet... A1 "starting with an a followed by any number of b's, then three 'c's and any number of a's.

A2 starting with an a followed by any number of 'c's, then three a's and any number of b's etc...

you cannot be both deterministic and simply have two paths from the same state to separate paths.

Remember! there is a deterministic solution to the above A1,A2 union languages as they are both regular!- however, simply joining the sub-graphs is not it!

AGAIN - notice (and this will be the 'style' that I will be adhering towards in solving / understanding our languages, and their properties - as mentioned- it is not the easiest solution that we use to represent our problem(s) it is rather the more sophisticated / interesting problem that is needed to emphasize the point. Our goal here is too understand the 'proof by construction' presented on pp 45 /46 (sipser)

this is where we rely on the construction supported by the suggestion of the catesian product

What in this case am i takin the cartesian product of ? it is of the states represented by individual graph structures

consider two minimal languages towards our example....

(for brevity lets keep these really simple - how about a one word language!) over the alphabet of {a,b,c} ... these languages have an individual word as there language

(why NOT KEEP THINGS SIMPLE FOLKS!) A1 abbbc A2 acbc

now what would our UNION of A1 U A2 language consider (of course you know this!!)

that is how #1 starts to look - you have to consider everything...

#2 is straightforward with my example as you have the same alphabet but lets say you had a second alphabet with e,f,g then you would have a,b,c,e,f,g

#3 this says my output state will be equivalent to the individual output states

insert (q1/q8 , 'a') in the left handside here.... (( in T((r1,r2),a) = (T(r1,a),T(r2,a)) )) ***** note .... i substituted 'T' for the greek transition symbol character!

this should take me to q2/q9 here!

more to come!

note that this can be easily modified to support intersection as noted in (sipser) text.......