Simpler description of pushout in $\mathsf{Set}$ and $\mathsf{Top}$ when one of the two maps is injective

Question

It is known that the pushout in $\mathsf{Top}$ has same underlying set as the pushout in $\mathsf{Set}$ (this is because the forgetful functor $\mathsf{Top}\to\mathsf{Set}$ is a left adjoint). Consider, either in $\mathsf{Set}$ or in $\mathsf{Top}$, a pushout diagram \begin{align*}\tag{1}\label{diag} \require{AMScd} \begin{CD} A @>{g}>> Y\\ @V{f}VV @VV{\overline{f}}V \\ X @>>{\overline{g}}> Z \end{CD} \end{align*} where $Z$ has underlying set $\frac{X\sqcup Y}{\mathcal{R}}$, and where the relation $\mathcal{R}$ on $X\sqcup Y$ equals $\mathcal{R}=R^e$, where $R$ is the relation $$ R=\{(f(a),g(a))\mid a\in A\}\subset(X\sqcup Y)^2, $$ and where we denote $(-)^e$ to the equivalence closure (defined to be the smallest equivalence relation on a set that contains some given relation).

The situation becomes more interesting if we suppose that at least one of $f$ or $g$ are injective. On one hand, if $f$ is injective we get that the map $\overline{f}$ from \eqref{diag} turns out to be injective as well (see this answer). On the other hand, if at least one of $f$ or $g$ are injective, then now the relation $\mathcal{R}$ can be also written as $$\tag{2}\label{eq} \mathcal{R}=R^e=\tilde{R}^{rs}, $$ where $\tilde{R}$ is the relation $$ \begin{align} \tilde{R}=R&\cup\{(f(a),f(b))\mid a,b\in A,\; g(a)=g(b)\}\\ &\cup\{(g(a),g(b))\mid a,b\in A,\; f(a)=f(b)\}, \end{align} $$ and where we denote $(-)^r$ and $(-)^s$, respectively, to the reflexive and symmetric closure (these closures are defined to be, respectively, the smallest reflexive or symmetric relation on a set that contains some given relation. It can be checked that the reflexive and symmetric closures commute, $(-)^{rs}=(-)^{sr}$). The explanation of where the equation \eqref{eq} comes from can be found on the comments of this answer.

Because of \eqref{eq}, I was wondering whether a more simple description of $\mathcal{R}$ is possible in the case that at least one of $f$ or $g$ are injective. You know, a simple characterization of the type “for $u,v\in X\sqcup Y$, we have $u\mathcal{R}v$ if and only if...”.

Answer 1

If $X$ is any set and $S\subset X\times X$ is any binary relation on $X$, then it is not difficult to verify that, for $x,x'\in X$, we have that $xS^{rs}x'$ if and only if $x=x'$ or $xSx'$ or $x'Sx$.

Suppose now $f$ is injective and let $u,v\in X\sqcup Y$ be such that $u\mathcal{R}v$ ($\Leftrightarrow u\tilde{R}^{rs}v$). If further $u,v\in Y$, then $u=v$ for $\overline{f}$ is injective. Suppose $u\neq v$. Then one and only one of the following holds (for $X$ and $Y$ are disjoint inside $X\sqcup Y$):

1. $u,v\in X$,
2. $u\in X$, $v\in Y$,
3. $u\in Y$, $v\in X$.

Since $u\tilde{R}^{rs}v$, this means that one and only one of the following holds:

1. $\exists a,b\in A:u=f(a),\; v=f(b),\; g(a)=g(b)$,
2. $\exists a\in A: u=f(a),\; v=g(a)$,
3. $\exists a\in A: u=g(a),\; v=f(a)$.

Putting all together, we conclude that, for $u,v\in X\sqcup Y$ and if $f$ is injective, we have that $u\mathcal{R}v$ if and only if at least one of the following holds:

1. $u=v$,
2. $\exists a,b\in A:u=f(a),\; v=f(b),\; g(a)=g(b)$,
3. $\exists a\in A: u=f(a),\; v=g(a)$,
4. $\exists a\in A: u=g(a),\; v=f(a)$.

This would constitute the ‘simple’ characterization of $\mathcal{R}$.

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As an addition, I will write this characterization in a particular case: the more usual topological construction of the ‘adjunction space’, where $A\subset X$, and we have the pushout diagram \begin{align*} \require{AMScd} \begin{CD} A @>{f}>> Y\\ @V{i}VV @VVV \\ X @>>> Z \end{CD} \end{align*} where $f$ is a continuous map and $i$ is the inclusion. This is usually denoted $Z=Y\cup_f X$.

Now the characterization reads: for $u,v\in X\sqcup Y$, we have $u\mathcal{R}v$ if and only if at least one of the following holds:

1. $u=v$,
2. $\exists a,b\in A: u=a,\;v=b,\;f(a)=f(b)$,
3. $\exists a\in A:u=a,\;v=f(a)$,
4. $\exists a\in A:u=f(a),\;v=a$.