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From here:

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I'm trying to verify that the isomorphism $\mathscr P(A)\cong H_2(A)$ is natural in $A$.

Let $f^{op}:A\to B$ be an arrow in $\mathbf{Set}^{op}$ (so $f:B\to A$ is an arrow in $\mathbf{Set}$). We need to verify that the diagram

$$ \require{AMScd} \begin{CD} \mathscr P(A) @>{\mathscr P(f^{op})}>> \mathscr P(B);\\ @VVV @VVV \\ H_2(A) @>{H_2(f^{op})}>> H_2(B); \end{CD}$$

commutes.

If we first go right and then down, then $U$ first gets sent to $f^{-1}(U)$ and then to the characteristic function $\chi_{f^{-1}(U)\subset B}$ of the subset $f^{-1}(U)\subset B$.

If we first go down and then right, then $U$ first gets sent to the characteristic function $\chi_{U\subset A}$ of $U\subset A$ and then to $\chi_{U\subset A}\circ f$.

One thing that got me confused is this: we have an arrow $A\to B$ (i.e., a function of sets), how is the arrow (function) $B\to A$ chosen? For example, if $A=\{\star,\ast\},B=\{\cdot\}$ and the functon $A\to B$ sends $\ast,\star\mapsto \cdot$, then we cannot construct a function $B\to A$ by sending $\cdot$ to both $\star,\ast$, so one has to make a choice.

And secondly, I don't quite see why the two compositions described above are the same.

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  • $\begingroup$ The arrow $A\to B$ coming from the opposite category is not a function of sets from $A\to B$, it is an arrow $B \to A$ in Sets, ie a function $B\to A$ already. $\endgroup$ – Ben Jul 1 '19 at 1:10
  • $\begingroup$ If you don’t see why the two compositions are the same you can try evaluating them at various elements in $B$ (Ones which map to $U$, or which dont). $\endgroup$ – Ben Jul 1 '19 at 1:20
  • $\begingroup$ @Ben Then if we have an arrow (a function) $h: B\to A$, then what exactly is the arrow (function) $h^{op}:A\to B$? $\endgroup$ – user634426 Jul 1 '19 at 1:20
  • $\begingroup$ Arrows $X\to Y$ are not always functions $X \to Y$ (even if $X,Y$ are sets). They are in the category of Sets, but not in Sets^op. In Sets^op, arrows $X\to Y$ are functions $Y \to X$. Given an arrow (function) $h: B \to A$, the arrow (not function) $h^{op}: A \to B$ is the function $h: B \to A$ (again). $\endgroup$ – Ben Jul 1 '19 at 2:31
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  1. The arrows in $\mathbf{Set}$ are functions. The arrows in $\mathbf{Set}^{op}$ are also functions, except we flip our labels for domain and codomain. So for example, there is a constant function $\{*,\star\}\to\{\bullet\}$ in $\mathbf{Set}$. This same function counts as an arrow $\{\bullet\}\to\{*,\star\}$ in $\mathbf{Set}^{op}$. In moving to the opposite category, we simply flip the labels for domain and codomain; the behavior is the same.

  2. Pick a function $f:B\to A$ in $\mathbf{Set}$. We can view it as an arrow $f^{op}:A\to B$ in $\mathbf{Set}^{op}$.

  3. Then $\mathscr{P}(f^{op}:A\to B)$ is a function from $\mathscr{P}(A)$ to $\mathscr{P}(B)$. It sends subsets of $A$ to their inverse image in $B$ under $f$. Note that because $f$ is a function from $B\to A$, this checks out.

  4. And $H_2(f^{op}:A\to B)$ is a function from $2^A$ to $2^B$. Given a function $A\to 2$, you can pre-compose with $f$ to get a function $B\to A \to 2$.

  5. The isomorphism $\mathscr{P}(A)\to H_2(A)$ sends each subset $U$ to its characteristic function $\chi(u) = \begin{cases}1, u\in U\\ 0, u\notin U\end{cases}$.

  6. It is natural because one way – right then down – we get the map that sends each subset $U$ of $A$ to the characteristic function of $f^{-1}(U)$. In other words, each subset $U\subseteq A$ gets sent to the map $\small(B\to 2)$ which assigns $1$ to every point $x\in B$ with $f(x)\in U$ and $0$ to every other point.

    The other way – down then right – $U$ gets sent to the characteristic function of $U$ pre-composed with $f$. In other words, this composition sends each point $x\in B$ to $f(x)$, then returns $1$ if $f(x)\in U$ and returns $0$ otherwise. This is the same function as before, so the isomorphism is natural.

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  • $\begingroup$ 1. But isn't an arrow $x\to y$ in $\mathbf {Set}^{op}$ a function with domain $x$ and codomain $y$ as well? If it is, then our arrow $\{\bullet\}\to\{\ast,\star\}$ must send $\bullet$ somewhere. $\endgroup$ – user634426 Jul 1 '19 at 2:19
  • $\begingroup$ @user634426 It isn't, actually. An arrow $A\rightarrow B$ in $\mathrm{Set}^{op}$ is exactly the same as a function from $B\rightarrow A$. That is, an arrow in $\mathrm{Set}^{op}$ with domain $A$ and codomain $B$ is the same as a function with domain $B$ and codomain $A$. In moving to the opposite category, nothing changes except what you label as domain and codomain. $\endgroup$ – user326210 Jul 1 '19 at 2:36
  • $\begingroup$ As an extreme example, in $\mathrm{Set}$, the empty set $\varnothing$ is the initial object. There is one, and only one, function from $\varnothing$ into each set. There are no functions into it except id. In $\mathrm{Set}^{op}$, the arrows are reversed so the dual is true: $\varnothing$ is terminal; there is exactly one arrow in $\mathrm{Set}^{op}$ with domain $A$ and codomain $\varnothing$. That arrow is the unique function $\varnothing \rightarrow A$ in $\mathrm{Set}$. $\endgroup$ – user326210 Jul 1 '19 at 2:40
  • $\begingroup$ If you like, in opposite categories, the definition of a particular function $f:A\rightarrow B$ remains the same; we just choose to write the arrows the other way: $f:A\leftarrow B$. $\endgroup$ – user326210 Jul 1 '19 at 2:45

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