# Question on inclusion of the lower horizontal map of a pushout diagram when the upper horizontal map is an inclusion.

Consider the following pushout diagram in the category $$\textbf {Top}$$ $$:$$

$$\require{AMScd} \begin{CD} A @>{\iota}>{\text {inclusion}}> B\\ @VVV @VV{}V\\ Y @>{f}>{}> X\end{CD}$$

$$\textbf {Question}$$ $$:$$ Is it true that the map $$f : Y \longrightarrow X$$ is also an inclusion? If so, how do I argue that?

• What is your definition of an inclusion? Injective map? Homeomorphism onto its image? May 14 at 19:52
• @Martin Brandenburg injective map. May 14 at 20:51

Since your definition of "inclusion" means "injective map", and the forgetful functor $$\mathbf{Top} \to \mathbf{Set}$$ preserves pushouts (in fact, all colimits), the question reduces to a pure set-theoretical one: The pushout of an injective map in $$\mathbf{Set}$$ is again injective. Although a direct proof, using the construction of the pushout, is possible (see the other answer), there is also a more elegant way to see this: First, it is easy to see that the result is true for injective maps of the form $$\emptyset \to X$$, since here the pushout with the unique map $$\emptyset \to Y$$ is just the inclusion into the coproduct $$Y \to X + Y$$. In all other cases, the injective map is a coretraction (aka section), and now we can use the following result, which holds in any category: If $$f : X \to Y$$ is a coretracton and $$g : X \to X'$$ is any morphism such that the pushout $$f' : X' \to X' \sqcup_X Y$$ exists, then $$f'$$ is a coretraction as well, and hence a monomorphism. In fact, choose $$r : Y \to X$$ with $$r \circ f = \mathrm{id}_X$$, then $$\mathrm{id}_{X'} : X' \to X'$$ and $$g \circ r : Y \to X'$$ agree on $$X$$ (i.e. $$\mathrm{id}_{X'} \circ g = g \circ r \circ f$$), hence induce a morphism $$h : X' \sqcup_X Y \to X'$$ with $$h \circ f' = \mathrm{id}_{X'}$$.
Here a proof that if $$\iota$$ is injective, then $$f$$ is injective as well. A direct computation, shows that $$X\cong (Y\amalg B)/{\sim}$$ where $$\sim$$ is the equivalence relation generated by: $$(0,\alpha(a))\sim(1,\iota(a))$$ for $$a\in A$$, where $$\alpha:A\to Y$$ and $$Y\amalg B=(\{0\}\times Y)\cup(\{1\}\times B)$$. Consequently, this reduces to show that for every $$y,y'\in Y$$ $$(0,y)\sim(0,y')\implies y=y'$$ Assume $$(0,y)\sim(0,y')$$. If $$y$$ doesn't belong to the image of $$\alpha$$, then clearly $$y=y'$$. Otherwise, there exists $$n\in\Bbb N$$ and $$a_i\in A$$ for $$0\leq i\leq n$$ such that $$y=\alpha(a_0)$$, $$y'=\alpha(a_n)$$ and $$(1,\iota(a_{2k}))=(1,\iota(a_{2k+1}))$$, $$(0,\alpha(a_{2k-1}))=(0,\alpha(a_{2k}))$$ for every $$k$$, that's: $$(0,\alpha(a_0))\sim(1,\iota(a_0))=(1,\iota(a_1))\sim(0,\alpha(a_1))=(0,\alpha(a_2))\sim\cdots\sim(0,\alpha(a_n))$$ Then $$a_{2k}=a_{2k+1}$$ for every $$k$$, hence $$\alpha(a_{2k})=\alpha(a_{2k+1})$$ and $$\alpha(a_{2k-1})=\alpha(a_{2k})$$, from which $$y=\alpha(a_0)=\alpha(a_n)=y'$$.