Pointed sets are equivalent to sets and partial functions but not isomorphic as categories

A partial function $A\rightharpoonup B$ is a function from a subset of $A$ to $B$, with the usual partial composition. Let $L(A)=A⨿\{\ast \}$ and extend a partial function by sending every undefined input and the new point to the new point. This defines $L:\mathbf{P}\mathbf{a}\mathbf{r}\to {\mathbf{S}\mathbf{e}\mathbf{t}}_{\ast }$.

Conversely, let $R(X,x_0)=X\setminus \{x_0\}$. A pointed map $f:(X,x_0)\to (Y,y_0)$ induces the partial function defined at $x\mathrm{\ne }{x}_0$ exactly when $f(x)\mathrm{\ne }{y}_0$, with value $f(x)$. This defines $R:{\mathbf{S}\mathbf{e}\mathbf{t}}_{\ast }\to \mathbf{P}\mathbf{a}\mathbf{r}$.

The empty set is the unique zero object of $\mathbf{P}\mathbf{a}\mathbf{r}$: if $Z$ were terminal, the empty partial function and each everywhere-defined map $\{\ast \}\to Z$ would force $Z=\mathrm{\varnothing }$. In ${\mathbf{S}\mathbf{e}\mathbf{t}}_{\ast }$ every pointed singleton is a zero object, so the distinct objects $(\{0\},0)$ and $(\{1\},1)$ are both zero objects.

Direct inspection of domains shows that both assignments preserve identities and partial composition. The canonical bijection $A\to RLA$ and the pointed bijection $(X\setminus \{x_0\})⨿\{\ast \}\to X$ that is inclusion on the first summand and sends $\ast$ to $x_0$ are natural. Hence $RL\cong 1_{\mathbf{P}\mathbf{a}\mathbf{r}}$ and $LR\cong 1_{{\mathbf{S}\mathbf{e}\mathbf{t}}_{\ast }}$.

Step 2.1 supplies the equivalence in [L1]. An isomorphism as in [L2] would biject objects and, together with its inverse, preserve and reflect the zero-object property, contradicting step 1.3. Thus these categories are equivalent but not isomorphic.