There is an interesting relation in which the action of adjoining a unit to an algebra and the one-point compactification come together:
Let $X$ be a non compact but locally compact Hausdorff space and let $C_0(X)$ denote the non-unital algebra of continuous functions on $X$ which vanish at infty (w.r.t. the sup-norm this is a C-*-algebra). We can adjoin a unit element $e$ to $C_0(X)$ which yields the unital C*-Algebra $\widetilde{C_0(X)}$. On the other side, we can compactify $X$ to $X^*$ and consider the C-*-algebra of continuous functions $C(X^*)$:
$$ X \to C_0(X) \to \widetilde{C_0(X)} \\ X \to X^* \to C(X^*) $$
Interestingly both paths have the same result modulo isomorphic equivalence, i.e. $$ \widetilde{ C_0(X) } \overset{\text{C*}}{\simeq} C(X^*). $$
Reading about this in Pedersen's Analysis Now, p.129:
If a Banach algebra $\mathfrak A$ has no unit, we may try to embed it isometrically into a larger unital Banach algebra $\mathfrak B$, in such a way that $\mathfrak A$ becomes an ideal in $\mathfrak B$, so large that every nonzero ideal of $\mathfrak B$ has a nonzero intersection with $\mathfrak A$ (an essential ideal). This process is the algebraic counterpart of the compactification of a topological space (1.7.2). For the classical Banach algebras there is often a natural way of adjoining a unit; but there is always an abstract procedure - the counterpart of the one-point compactification.
Another place in the web where I found a similar statement was in this post:
Adjoining a unit corresponds to passing to the one-point compactification.
I am wondering if this "phenomenon" can be described in terms of category theory (as e.g. in the case of the Banach-Stone theorem) and if there are further examples of this interplay between the one-point compactification and adjoining a unit.
In particular, I am interested in which sense one should understand the terms counterpart or corresponds in the two cited sources.
For the sake of completeness, here is my proof of the mentioned equivalence.
Let $X$ be a locally compact Hausdorff space and $X^* = X \cup \{\infty\}$ its one-point compactification. Furthermore, let $(C_0(X), \|\cdot\|_\infty)$ be the C*-Algebra of continuous functions vanishing at infinity and $(C(X^*)), \|.\|_\infty)$ be the C*-algebra of continuous functions on $X^*$. Then, $$ \widetilde{C_0(X)} \simeq C(X^*), $$ where $\widetilde{\mathcal A}$ denotes the unification of a non-unital algebra $\mathcal A$.
Proof. Consider the mapping $$ \varphi\colon \widetilde{C_0(X)} \to C(X^*), \quad (f, \alpha) \mapsto \overline f + \alpha := \begin{cases} f(x) + \alpha, \quad & x \in X, \\ \alpha, \quad & x = \infty. \end{cases} $$ We want to show $$ \|\varphi(f, \alpha)\|_\infty = \|(f,\alpha)\|_{\widetilde{C_0(X)}}. $$ For this, we use the following
Lemma. For $f \in C(X^*)$, we have $$ s_1 := \sup_{x \in X} |f| = \sup_{x \in X^*} |f| =: s_2. $$
Proof. Let $x_\alpha$ be a net in $X$ with $\lim_\alpha x_\alpha = \infty$. Then, $f(x_\alpha) \leq s_1$ and thus $f(\infty) \leq s_1$. This gives us $s_2 \leq s_1$. The other inequality is obvious. $\square$
On the one hand, we have $$ \|\varphi(f,\alpha)\|_\infty = \sup_{x \in X^*} |\;\overline f(x) + \alpha\;| = \sup_{x \in X} |\;\overline f(x) + \alpha\;| = \sup_{x \in X} |\;f(x) + \alpha\;| = \|f + \alpha \|_\infty. $$ On the other hand, we have $$ \|(f, \alpha)\|_{\widetilde{C_0(X)}} = \sup_{g \in C_0(X), \|g\| \leq 1} \|fg + \alpha g\|_\infty \leq \sup_{g \in C_0(X), \|g\|\leq 1} \|g\|_\infty \|f + \alpha\|_\infty \leq \|f + \alpha\|_\infty = \|\varphi(f, \alpha)\|_\infty . $$
Now, let $(g_i)_{i \in I}$ be an approximate unit in $C_0(X)$. Then, $$ \|f g_i + \alpha g_i\|_\infty \to \|f + \alpha\|_\infty $$ which in turn implies $$ \|(f, \alpha ) \|_{\widetilde{C_0(X)}} = \|\varphi(f, \alpha)\|_\infty. $$ This proves the claim. $\square$
Feel free to edit this proof or leave a comment if I made a mistake.
