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.
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.