Per your comment, I think you misunderstood what Munkres is trying to say.
If Y is subspace of X, a separation of Y is a pair of disjoint nonempty sets A and B whose union is Y, neither of which contains a limit point of the other. The space Y is connected if there exists no separation of Y.
I read it to mean two definitions are given. Firstly, that he defines what it means to have a separation of a subspace $Y$ inside of the space $X$. Then he defines a space $Y$ as connected if it cannot be separated within itself. ($Y$ is trivially a subspace of itself, so the first definition of a separation can be used.)
Now, in the example I gave where $X = \{a,b,c\}$, $Y = \{a,c\}$, with topology on $X$ generated by $\{a,b\}, \{b,c\}$, the subspace $Y$ is not a connected set in $X$, as it is not a connected space in its subspace topology. But the space $X$ is connected, so the connected component of $\{a\}$ in $X$ is the whole space. (Whereas the connected component in $Y$ is itself.)
This shouldn't be so strange if you consider a more intuitive example: Let $X$ be the open interval $(0,1)$, and $Y$ be the subset $(0,1/4)\cup (3/4,1)$. Then $Y$ is not connected. The connected component containing the point $1/8$ in $X$ is the whole space, whereas the connected component when considered in $Y$ is just the interval $(0,1/4)$.
In other words, the connected component of a point in $X$ is a subspace $Y$ such that $Y$ is connected in the subspace topology and such that $Y$ and $X\setminus Y$ are both open. (And $Y$ of course contains the point in question.)
The unitalization approach can be made to work.
Let $C_K = \{ (r,s) \in R \times R \mid r-s \in K \}$ be the congruence defined by an ideal $K$.
Then, we have three maps defined on $S = C_I \otimes_R C_J$ :
- $\pi_0 : S \to C_I$ induced by the first projection $C_J \to R$ and the identity on $C_I$
- $\pi_1 : S \to C_J$ induced by the first projection $C_I \to R$ and the identity on $C_J$
- $\mu : S \to R \times R$ induced by the inclusions $C_I \to R\times R$ and $C_J \to R \times R$.
Letting $\Delta \subseteq R \times R$ be the image of the diagonal, define $T = \{ x \in S \mid \pi_0(x) \in \Delta \wedge \pi_1(x) \in \Delta \}$. I claim that $\mu(T) = C_{IJ}$.
On the level of $R$ modules, we have isomorphisms $C_K \cong R \oplus K$, such as $(r,s) \mapsto (r, s-r)$, and so we have
$$ S \cong R \oplus I \oplus J \oplus (I \otimes_R J) $$
In this form, the maps $\pi_i$ become projections onto the relevant summands, so $T$ is precisely the submodule $R \oplus (I \otimes_R J)$, so we've eliminated the $I$ and $J$ summands you were having trouble with.
By splitting the $R$-module maps, $T$ is genenerated as an $R$-module by elements of the form
$$ (r, r+i) \otimes (s, s+j) - (0,i) \otimes (s,s) - (r,r) \otimes (j,0) $$
and applying $\mu$ to such a thing gives the element $(rs, rs + ij)$, and now it's easy to see that $\mu(T) = C_{IJ}$ as claimed.
Best Answer
Inclusions of subspaces are precisely the regular monomorphisms, and projections of quotients are precisely the regular epimorphisms.