Can anyone give me a "hands on" proof of why the pushout of an injective function (along any other function) is again an injective function?
I know sets form a topos and toposes are adhesive, but is there a more elementary proof in the case of sets, that show better what is "really going on"?
The pushout of $Y \leftarrow^g Z \to^f X$ in $\mathsf{Set}$ can be constructed as follows: it is the quotient of $X \sqcup Y$ by the equivalence relation generated by $g(z) \sim f(z)$. That is, $x,x' \in X$ are equivalent if and only if there is a "zig-zag" $$ x = f(z_0) \sim g(z_0) = g(z_1) \sim f(z_1) = f(z_2) \sim \cdots \sim f(z_n) = x'. $$ Now if $g$ is injective, there are no non-trivial such zig-zags, because $g(z_0) = g(z_1) \iff z_0 = z_1$. Thus, $x \sim x'$ if and only if $x = x'$. Hence the mapping from $X$ to the pushout -- which is the inclusion in $X \sqcup Y$ followed by the quotient mapping -- is injective.