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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?

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    $\begingroup$ What is your definition of an inclusion? Injective map? Homeomorphism onto its image? $\endgroup$ May 14 at 19:52
  • $\begingroup$ @Martin Brandenburg injective map. $\endgroup$
    – Anacardium
    May 14 at 20:51
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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 $h \circ f' = \mathrm{id}_{X'}$.

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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 $y=\alpha(a_0)=\alpha(a_n)=y'$.

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