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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?

Wolgwang
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wormram
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  • The pushout has a universal property; you need to use that universal property, or you have no hope of proving this. Otherwise, you could consider the one-point space and the maps from $X$ and $Y$ to the one-point space, which also satisfy the diagram commutativity but where the composition is not injective. – Arturo Magidin Nov 13 '20 at 19:41
  • Sorry, fixed the typos. I know the universal property, but I'm not sure how to apply it here. Won't I need a third space $Z$ with maps $\phi_1:X\to Z$ and $\phi_2:Y\to Z$ to use it? – wormram Nov 13 '20 at 19:45
  • Yes; you need to find a cleverly defined $Z$, $\phi_1$ and $\phi_2$ (with $\phi_1 \circ f = \phi_2\circ g$) to deduce injectivity. The point is that if you just took $Z={\star}$, and defined $\phi_1$ and $\phi_2$ as the unique maps, then you would have all the properties you are using so far, but you would not be able to conclude that the map $Y\to Z$ is one-to-one... because it isn't. So you need to use something about the pushout that goes beyond "the square commute". And the only thing that the pushout has that goes beyond that is the universal property. – Arturo Magidin Nov 13 '20 at 19:47
  • Of course, alternatively you can avoid the universal property and just deal with the construction. Note that elements outside the image of $A$ in $Y$ are not identified with anyone, and no point outside the image of $A$ is identified with a point inside the image of $A$. So you'll just have to consider two points in the image of $A$ as being potentially identified. – Arturo Magidin Nov 13 '20 at 20:08
  • The definition of ~ is circular with bad quantification. – William Elliot Nov 13 '20 at 22:25
  • @ArturoMagidin If you mean by "using the universal property of the pushout" that one can give a proof on an abstract categorical level, you may be right for some properties. But some properties are specific for a category, i.e. require the explicit construction of pushouts. For example, pushouts of embeddings in $\mathbf {Top}$ are embeddings, but it seems that there is no purely categorical proof of this fact. – Paul Frost Nov 15 '20 at 01:03
  • @PaulFrost: I meant to choose specific topological spaces and then invoke the universal property, so a kind of “hybrid” approach in which we use topology to get the “well-chosen $Z$”, and then the category theory to get the maps into that well-chosen $Z$. You are correct that there is no purely categorical proof of this, because the canonical maps into the pushout are not always monomorphisms. – Arturo Magidin Nov 15 '20 at 01:05
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    @ArturoMagidin Okay, we agree! – Paul Frost Nov 15 '20 at 01:06

1 Answers1

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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.

Arturo Magidin
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