Let $f:A\to X$ be a continuous injective map of spaces. Let $g:A\to Y$ be continuous. Prove that the map $i_Y:Y\to X\cup_A Y$ is injective. Here, $X\cup_A Y:=X\coprod Y/\sim$, where $\sim$ is the finest equivalence relation such that $f(a)\sim g(a)$. This is called the pushout of $X$ and $Y$ along $A$. Also, the map $i_Y$ is defined as the composition of the inclusion $X\to X\coprod Y$ with the quotient map $X\coprod Y\to X\coprod Y/\sim$.
I'm new to the idea of pushouts, so I'm having a hard time getting started here. I know that we have the "diagram commutativity" relation $i_X\circ f=i_Y\circ g$ where the definition of $i_X$ is analogous to the definition of $i_Y$, but I'm not sure how to use this. My first thought was to just jump in and try to prove this using the definition of injective as follows.
Pick $y_1,y_2\in Y$ such that $i_Y(y_1)=i_Y(y_2)$. A first case I tried considering was when $y_1=g(a_1)$ and $y_2=g(a_2)$ for some $a_1,a_2\in A$. Then we have (using $[-]$ to denote eq. class),
$$
[f(a_2)]=[g(a_2)]=i_Y(y_2)=i_Y(y_1)=[g(a_1)]=[f(a_1)],
$$
and then the injectivity of $f$ implies $a_1=a_2$? I'm not sure here. Also, what about the cases when $y_1$ and $y_2$ are not in the image of $g$?
Any suggestions? Am I approaching this right?
Best Answer
One nice thing about the category of all topological spaces is that you can make your favorite set-theoretic function continuous by either making the codomain have the indiscrete topology, or make the domain have the discrete topology. This means that you have a lot of liberty for choosing when you have a property that guarantees the existence of a map whenever “there exists a topological space $Z$ and continuous functions with properties
blah.” This is a consequence of the fact that the underlying set functor from $\mathsf{Top}$ to $\mathsf{Set}$ has both a left and a right adjoint, given precisely by the discrete topology functor and the indiscrete topology functor, respectively.So... for this problem. Since we want maps into a “well-chosen $Z$”, we probably want to endow $Z$ with the indiscrete topology; then we won’t have to worry about continuity and can just define the maps any way we want, to make sure they satisfy the relevant properties. Here goes.
If $Y$ is empty, then $A$ is empty and there is nothing to do: any map with empty domain is injective. So assume $Y$ is not empty and let $y_0\in Y$.
Let $Z$ be the topological space with underlying set $Y$ and the indiscrete topology. Define the map $\phi_2\colon Y\to Z$ to be the identity on underlying sets. Define the map $\phi_1\colon X\to Z$ as follows:
If $x\in X$ and there exists $a\in A$ with $x=f(a)$, then let $\phi_1(x) = \phi_2(g(a))$ (really just $g(a)$, since the underlying set of $Z$ is the same as the underlying set of $Y$). This is well defined, because $f$ is injective. If $x\notin f(A)$, then let $\phi_2(x)=y_0$.
Since any map into an indiscrete space is continuous, both $\phi_1$ and $\phi_2$ are continuous. Moreover, if $a\in A$, then $\phi_1(f(a)) = \phi_2(g(a))$ by definition. Thus, by the universal property of the pushout, there exists a unique continuous map $\Phi\colon X\cup_A Y\to Z$ such that $\phi_1=\Phi\circ i_X$ and $\phi_2=\Phi\circ i_Y$. But $\phi_2$ is a bijection, so $\Phi\circ i_Y$ is a bijection. Therefore, $\Phi$ is surjective and $i_Y$ is injective. This shows that $i_Y$ is injective.