The Pushout of $X \leftarrow Z\rightarrow Y$ with $f:Z\rightarrow X$ and $g:Z\rightarrow Y$ in $\mathbf{Top}$ exists and is given by $X\coprod Y/\sim,$ where "$\sim$ is the equivalence relation generated by $i_X f(z)\sim i_Y g(z)$ for all $z\in Z$".
What does this last sentence mean explicitly? Thanks.
If $R$ is a relation on a set $X$, then the generated equivalence relation is by definition the smallest equivalence relation $A$ on $X$ such that $R \subseteq A$. It always exists (take the intersection of all equivalence relations containing $R$). One can describe it explicitly as follows: First one takes the reflexive closure: $(x,y) \in R_{\mathrm{ref}}$ iff $x=y$ or $(x,y) \in R$. Then one takes the symmetric closure: $(x,y) \in R_{\mathrm{ref},\mathrm{sym}}$ iff $(x,y) \in R_{\mathrm{ref}}$ or $(y,x) \in R_{\mathrm{ref}}$. Finally, one takes the transitive closure: $(x,y) \in R_{\mathrm{ref},\mathrm{sym},\mathrm{trans}}$ iff there is a finite sequence $x=x_0,\dotsc,x_n=y$ such that $(x_i,x_{i+1}) \in R_{\mathrm{ref},\mathrm{sym}}$. This is nothing else than a zig-zag path through the graph representing $R$ (draw it!).
The equivalence relation generated by $i_X f(z) \sim i_Y g(z)$ for all $z \in Z$ actually means the smallest equivalence relation which contains all $(i_X f(z),i_Y g(z))$ for $z \in Z$.