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I am supposed to find $k \in \mathbb{N}$ and attaching maps $\varphi_1,\varphi_2:S^k \rightarrow S^1$, such that there are two pushouts

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} S^{k} & \ra{\hspace{0.35cm} \varphi_1 \hspace{0.35cm}} & S^1 \\ \da{inc} & & \da{inc}\\ D^{k+1} & \ras{\hspace{9mm}} & D^2\\ \end{array} \\ \begin{array}{c} S^{k} & \ra{\hspace{0.35cm} \varphi_2 \hspace{0.35cm}} & S^1 \\ \da{inc} & & \da{inc}\\ D^{k+1} & \ras{\hspace{9mm}} & D^2\\ \end{array} $$

where '$inc$' denotes the set-inclusion in all cases, and $\varphi_1,\varphi_2$ are 'not isomorphic', i.e. do not differ by precomposition with a homeomorphism .

I feel like this is not possible and can't (yet) wrap my head around this... It's not possible for $k \lt 1$, since we couldn't get any 2-cells that way, and for $k=1$, I can't wrap my head around $\varphi_i$ being anything but the identity (because it seems like $\varphi_i \ne id$ won't glue $S^1$ to $D^2$ in such a way as to obtain $D^2$ as a pushout...

I thought of $k=2$, but only came up with projecting $S^2$ to $D^2$, then shrinking $D^2$ to an interval, then gluing the ends together - however, it seems like $D^3$ gets 'lost' in this case.

Last but not least, I feel like $k>2$ is not worth thinking of here.

Can someone give me a hint?

Alternatively, the counterexample could be loosened to allow

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \bigsqcup_{i \in I} S^{k} & \ra{\hspace{0.35cm} \varphi_1 \hspace{0.35cm}} & S^1 \\ \da{inc} & & \da{inc}\\ \bigsqcup_{i \in I} D^{k+1} & \ras{\hspace{9mm}} & D^2\\ \end{array} \\ \begin{array}{c} \bigsqcup_{i \in I} S^{k} & \ra{\hspace{0.35cm} \varphi_2 \hspace{0.35cm}} & S^1 \\ \da{inc} & & \da{inc}\\ \bigsqcup_{i \in I} D^{k+1} & \ras{\hspace{9mm}} & D^2\\ \end{array} $$ (note that the index sets are the same), if that helps.

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  • $\begingroup$ In your very first line, should $\varphi_1, \varphi_2$ have $S^k$ as their domain rather than $S^1$? And what, exactly, was the problem with $k = 1$? What do you know about maps from $S^1$ to $S^1$? And what does "isomorphic" in the title mean? Homeomorphic as CW complexes? Homotopy equivalent? Something else? $\endgroup$ Commented Apr 17, 2017 at 11:02
  • $\begingroup$ @JohnHughes Yes, and I have added an explanatory remark, thank you. What do I know about maps from $S^1$ to $S^1$? They correspond to loops into $S^1$ over some point (?). $\endgroup$ Commented Apr 17, 2017 at 11:42
  • $\begingroup$ @JohnHughes ... and if the attaching map is not the identity for $k=1$, then how do I obtain $D^2$ and an inclusion on the right side? As I wrote, it just seems completely counterintuitive to me, that glueing $S^1$ to $D^2$ with the inclusion on the right side with anything but the identity would be possible... $\endgroup$ Commented Apr 17, 2017 at 11:45
  • $\begingroup$ The bottom map need not be the identity. (Indeed, for $k > 1$, saying it's the identity doesn't even make sense). $\endgroup$ Commented Apr 17, 2017 at 11:52
  • $\begingroup$ @JohnHughes Of course it doesn't, but you asked "And what, exactly, was the problem with $k=1$?", to which I replied. And I wasn't talking about the bottom map... we are talking about the top map(s), which, since the inclusion is fixed on the left, determine the pushout square (and hence the bottom map) up to homeomorphy, so why would I be asking about the bottom map? ... $\endgroup$ Commented Apr 17, 2017 at 11:58

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Hint: Let me ask you a simpler question: Can you fill in the following diagram... $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} S^{1} & \ra{\hspace{0.35cm} \varphi_1 \hspace{0.35cm}} & S^1 \\ \da{id} & & \da{id}\\ S^{1} & \ras{\hspace{9mm}} & S^1\\ \end{array} $$

with the map $\varphi_1$ being anything other than the identity? If you coud, then you could

(1) replace the bottom right space with $D^2$ (the right-hand map becomes inclusion)

(2) replace the bottom left space with $D^2$ by extending the bottom map "radially" over the rest of the disk.

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  • $\begingroup$ Hmm... well, sure, if the lower map is $\varphi_1$, if that is what you want to hint at, and there can't be any other way. As far as the extension is concerned, I suppose you were thinking of something like $r\phi \mapsto r \varphi_1(\phi)$ (where I have sluggishly 'identified' $e^{(i \phi)} = \phi$ as an argument for \varphi_1). However, do you know, that the result will actually be a pushout square, though, and not a mere commutative square? $\endgroup$ Commented Apr 17, 2017 at 12:14
  • $\begingroup$ Well...that might take some proving. That's why this is a homework problem. :) So now the question is "What could $\varphi_1$ possibly be?" And the answer's not "the identity" (CERTAINLY not for $k > 1$!), so what else might be possible? $\endgroup$ Commented Apr 17, 2017 at 12:37
  • $\begingroup$ Ahhhhhhhh..... :D complex conjugation? What held me up was that I only thought of the antipodal map, but that would have yielded the real projective plane (not the same, since the homology groups are not all identical). Complex conjugation seems to make sense, when I try to visualize the pushout... $\endgroup$ Commented Apr 17, 2017 at 12:57
  • $\begingroup$ The antipodal map would have yielded a sphere. But even if it had yielded a projective plane, what does "not the same, since homology groups..." mean? Can you say what's not the same as what, and why it was supposed to be? $\endgroup$ Commented Apr 17, 2017 at 13:06
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    $\begingroup$ I just didn't want to claim that every degree-1 map was injective...so I noted that they were all "almost" injective, i.e., up to homotopy. I think I've said as much as I'm going to say at this point. I wish you the best of luck. $\endgroup$ Commented Apr 17, 2017 at 17:29

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