# Monic arrows as subobjects

I've been reading "Topoi: The Categorial Analysis of Logic" by Robert Golblatt. It's an introduction to Category theory to me, and the exposition has been wonderful so far. I've reached the part where he introduces subobjects. Here's where I get stuck

Exercise: In Set, $\mathrm{Sub}(D) \cong \mathcal{P}(D)$.

Here, $\mathrm{Sub}(D) = \{[f]: f \text{ is monic with }\mathrm{cod} \ f = d\}$, and $\mathcal{P}$ is the powerset, i.e. $\mathcal{P}(D) = \{A : A \text{ is a subset of } D\}$.

I tried proving the exercise by letting the monic arrows be inclusion functions, but I ran into a problem. Suppose $D = \{1,2,3,4,5,6\}$. I could have sets $A = \{1,2,3\}$ and $B = \{4,5,6\}$. Okay, map those sets to inclusion functions $inc_A : A \to D, inc_B : B\to D$, where inclusion is the identity function restricted to that set.

Here's where I run into a problem. If I let $h = x + 3$, then $inc_B \circ h = inc_A$, and similarly for $inc_B$. Then the subobjects $inc_A \simeq inc_B$. So they're part of the same equivalence class. But they're different subobjects! I feel like this flattens all the functions into any equal-cardinality bijection.

I'm pretty stuck. Any pointers? You don't have to solve the exercise, just small hints or maybe show where I made a mistake.

• There is indeed a function $\mathcal{P}(D) \to \operatorname{Sub}(D)$ that sends each subset $S$ to the inclusion $S \to D$ (this is basically the starting point you described). The trick to the exercise is to write down its inverse; the function $\operatorname{Sub}(D) \to \mathcal{P}(D)$, and show that it really is an inverse. The 'obvious' function defined on monics with codomain $D$ respects equivalence.
– user14972
Commented Mar 11, 2017 at 0:13

The issue is that $inc_B \circ h \neq inc_A$. $inc_A$ maps $x \mapsto x$, but $inc_B \circ h$ maps $x \mapsto x + 3$. So they aren't actually equal. I'm going to think about this again...