Are equivalence classes of subobjects of some X in Set just equivalence classes of subsets of X with a specific cardinality?

Question

I'm trying to understand what would be the subobjects of $\{0, 1\}$.

Would they be $\{\emptyset, \{0\}, \{1\}, \{0, 1\} \}$?

Or are $\{0\}$ and $\{1\}$ somehow identified together? Because I can map from 0 to 1 and backwards - which tells me they're a part of the same equivalence class.

And what's stopping me from picking $\{7\}$ as a subobject, based on this definition? There was a similar answer from a while ago, but I fail to see how it answers the questions I'm posing above.

Answer 1

I preassume that you are looking at set $\{0,1\}$ as an object of the category of $\mathbf{Sets}$ here.

Object $\{0,1\}$ has $4$ subobjects.

Each of them is a class of of injective functions that all have $\{0,1\}$ as codomain.

• If $m$ denotes the unique arrow $\varnothing\to\varnothing$ then $\{m\}$ is a subobject of $\{0,1\}$.

• The class of functions $\{a\}\to\{0,1\}$ where $a$ is sent to $0$ is a subobject of $\{0,1\}$.

• The class of functions $\{a\}\to\{0,1\}$ where $a$ is sent to $1$ is a subobject of $\{0,1\}$.

• The class of injective functions $\{a,b\}\to\{0,1\}$ with $a\neq b$ is a subobject of $\{0,1\}$.

Answer 2

In $Set$, at least it has exactly the four subobjects you wrote up.

Note that a subobject of some object $X$ is not only a mere object $A$ in the same category, but it is understood to be together with a monomorphism $i:A\hookrightarrow X$ which plays the role of the inclusion.
Said that, indeed nothing prevents us to take $\{7\}$ as a subobject of $\{0,1\}$, moreover there are exactly two ways to do that: $i$ either sends $7\mapsto 0$, or $7\mapsto 1$.
So the first one represents the same subobject as $\{0\}\hookrightarrow\{0,1\}$, while the second one represents the same subobject as $\{1\}\hookrightarrow\{0,1\}$, but these two are distinct.

Answer 3

A subobject of an object of a category is not an object of the same category, it is an equivalence class of monomorphisms to this object (see definition from your link). So $\varnothing$, $\{0\}$, $\{1\}$, $\{0,1\}$ are not subobjects of $\{0,1\}$, they are its subsets. But for every set $X$ the power set $\mathcal{P}(X)$ is isomorphic to the set of subobjects of $X$ in the category of sets (and they are isomorphic as ordered sets). For example, the subset $\{0\}\subset\{0,1\}$ corresponds to the subobject $[i_{\{0\}}]$, where $i_{\{0\}}\colon\{0\}\to\{0,1\}$ is the canonical inclusion of subset and $[-]$ is taking of equivalence class of a monomorphism (as in the definition by the link). It is easy to see that if $X$, $Y$, $Z$ are sets and $f\colon Y\to X$ and $g\colon Z\to X$ are monomorphisms (injections), then $[f]=[g]$ as subobjects of $X$ in the category of sets if and only if their set-theoretic images are equal: $f(Y)=g(Z)$. So, for example, if $i_{\{0\}}\colon\{0\}\to\{0,1\}$ and $i_{\{1\}}\colon\{1\}\to\{0,1\}$ are canonical inclusions, then $[i_{\{0\}}]$ and $[i_{\{1\}}]$ are not equal as subobjects of $\{0,1\}$ in the category of sets.