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Consider the pushout diagram in the category Top $:$

$$\require{AMScd} \begin{CD} A @>{f}>{\text {inclusion}}> B\\ @V{g}VV @VV{i_1}V\\ Y @>{i_2}>{}> X\end{CD}$$ Suppose that $(B,A)$ has HEP. Show that $(X,Y)$ also has HEP.

$\textbf {My attempt}$ $:$ We first note that $i_2$ is an inclusion since so is $f.$ Now since $(B,A)$ has HEP it follows that there exists a retraction $r : B \times I \longrightarrow (B \times \{0\}) \cup (A \times I).$ So in order to show that $(X,Y)$ has HEP we need to construct a retraction $s : X \times I \longrightarrow (X \times \{0\}) \cup (Y \times I).$ Now the above diagram of pushout induces a pushout of the following form $:$

$$\require{AMScd} \begin{CD} A \times I @>{f \times \text {id}}>{\text {inclusion}}> B \times I \\ @V{g \times \text {id}}VV @VV{i_1 \times \text {id}}V\\ Y \times I @>{i_2 \times \text {id}}>{\text {inclusion}}> X \times I \end{CD}$$

So by virtue of universal property of pushout, in order to get a map from $s : X \times I \longrightarrow (X \times \{0\}) \cup (Y \times I)$ it is enough to construct maps $j_1 :B \times I \longrightarrow (X \times \{0\}) \cup (Y \times I)$ and $j_2 : Y \times I \longrightarrow (X \times \{0\} \cup (Y \times I)$ such that $j_1 \circ (f \times \text {id}) = j_2 \circ (g \times \text {id}).$ Consider the map $\varphi : (B \times \{0\}) \cup (A \times I) \longrightarrow (X \times \{0\}) \cup (Y \times I)$ defined by $$\varphi(z)= \begin{cases} (i_1(b),0), & \text {if}\ z = (b,0),\ \text {for some}\ b \in B \\ (g(a),t), & \text {if}\ z = (a,t),\ \text {for some}\ a \in A\ \text {and}\ t \in I \end{cases}$$

Define $j_1 : = \varphi \circ r$ and $j_2 : = \text {id}_{Y \times I}.$ Then it is easy to see that $j_1 \circ (f \times \text {id}) = j_2 \circ (g \times \text {id}).$ So by universal property of pushout there exists a unique map $s : X \times I \longrightarrow (X \times \{0\}) \cup (Y \times I)$ such that the following holds $:$

$(1)$ $s \circ (i_1 \times \text {id}) = j_1$ and

$(2)$ $s \circ (i_2 \times \text {id}) = j_2.$

Now the second equality gives us $s \big\rvert_{Y \times I} = \text {id}_{Y \times I}.$ Whereas the first equality gives us $s \big\rvert_{i_1(B) \times \{0\}} = \text {id}_{i_1(B) \times \{0\}}.$ Instead we need $s \big \rvert_{X \times \{0\}} = \text {id}_{X \times \{0\}}$ because we are willing to show that $s$ is a retraction.

I got stuck at this stage. If $i_1$ is a surjection then we are through.If not, I don't know how to proceed further. Would anybody kindly help me in this regard? A small hint will be warmly appreciated at this stage.

Thanks for your time.

Anacardium
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  • Read this and then answer you own question. – Paul Frost May 15 '21 at 22:20
  • @Paul Frost I only need to convert $i_1 (B) \times {0}$ to $X \times {0}.$ For that a small hint will be enough, I guess. I don't have enough time to read your entire post. I will figure it out on my own then. – Anacardium May 16 '21 at 03:13
  • The proof is almost complete because $X \times {0} = (Y \times {0}) \cup_{g \times {0}} (B \times {0}).$ So in order to show that $s \big\rvert_{X \times {0}} = \text {id}{X \times {0}}$ we need only to show two things $:$ $$\begin{align*} (1)\ s \big\rvert{i_1(B) \times {0}} & = \text {id}{i_1(B) \times {0}},\ \text {and}\ (2)\ s \big\rvert{Y \times {0}} & = \text {id}_{Y \times {0}} \end{align*}$$ Fortunately I have both the things at my disposal and thus the proof is complete. @Paul Frost done it! – Anacardium May 19 '21 at 06:51

1 Answers1

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The proof is almost complete because $$X \times \{0\} = (Y \times \{0\}) \cup_{g \times \{0\}} (B \times \{0\})$$ by thinking of $X \times \{0\}$ as a pushout induced by the first pushout diagram since $\{0\}$ is locally compact. Now after identification the copy of $B$ sitting inside $X$ is precisely $i_1(B)$ and the copy of $Y$ sitting inside $X$ is $Y$ itself since $Y$ is embedded inside $X.$ So in order to show that $s \big\rvert_{X \times \{0\}} = \text {id}_{X \times \{0\}}$ we need only to show two things $:$

$(1)\ s \big\rvert_{i_1(B) \times \{0\}} = \text {id}_{i_1(B) \times \{0\}}$

$(2)\ s \big\rvert_{Y \times \{0\}} = \text {id}_{Y \times \{0\}}$

Fortunately we have both the things at our disposal and thus the proof is complete.

Anacardium
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