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Suppose I take the $0$-skeleton $X_0$ to have a single inhabitant $base$.

Let $S_1$ have a single point $base'$, with attaching map $f(base', -)=base$. If I construct the pushout to the 1-skeleton $X_1$, as below, $$ \require{AMScd} \begin{CD} S_1 \times \mathbb{S}^0 @>{f}>> X_0;\\ @VVV @VVV \\ \mathbf{1} @>>> X_1; \end{CD} $$ I end up with $X_1$ defined by

  • $inl: \mathbf{1} \rightarrow X_1$
  • $inr: X_0 \rightarrow X_1 $
  • for each $c:S_1 \times \mathbb{S}^0$, a path $inl(\star) = inr(base)$

This is the interval type though! Where have I gone wrong in my construction?

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Nothing wrong in your construction. You don't get the interval since there is no way even to prove that your two paths are equal. For clarity, note that $\mathbb{S}^1$ as a pushout is simply the pushout of the following span

$$ \begin{CD} \mathbb{S^0} @>>> \mathbf{1} \\ @VVV \\ \mathbf{1} \end{CD} $$, where $\mathbf{1}$ is the unit type and $\mathbb{S}^0$ is the type of booleans (or the coproduct $\mathbf{1}\coprod\mathbf{1}$ if you wish).

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  • $\begingroup$ Thanks! My mistake was that I thought $\mathbb{S}^0$ was a singleton, rather than having two inhabitants. $\endgroup$ – Dr. John A Zoidberg Jun 28 '17 at 16:47
  • $\begingroup$ How would I construct the interval from a pushout then? $\endgroup$ – Dr. John A Zoidberg Jun 28 '17 at 16:50
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    $\begingroup$ @Dr.JohnAZoidberg The interval is equivalent to $1$, so it's not clear that that's a meaningful question. For instance, it's a pushout of $1\leftarrow 1\to 1$... $\endgroup$ – Kevin Carlson Jun 29 '17 at 1:41
  • $\begingroup$ @Dr.JohnAZoidberg, you are welcome! $\endgroup$ – Anthony Bordg Jun 29 '17 at 12:05
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I don't see why you say this is the interval type. There are two points in $S_0\times \mathbb{S}^0$, so you get to paths between the two given points, which is indeed a way of describing the circle.

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  • $\begingroup$ Is it possible to prove the equivalence of a circle defined by two points and two paths , and the circle defined by $base$ and $loop$? $\endgroup$ – Dr. John A Zoidberg Jun 28 '17 at 18:10
  • $\begingroup$ @Dr.JohnAZoidberg, yes it is possible and I will be happy to give further details if you open a new question since it is not convenient to answer questions in the comments section. $\endgroup$ – Anthony Bordg Jun 29 '17 at 12:02

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