Why is this pushout equal to $S^2$?

Question

Let $i:S^1 \rightarrow D^2$ be the inclusion and $p:S^1 \rightarrow \{∗\}$ be the projection. Construct continuous mappings $f:D^2 \rightarrow S^2$ and $j:\{∗\} \rightarrow S^2$ such that the following diagram is a pushout:

This is an old exercise of this semester. However, I didn't do it back then and now I want to do it for exam preparation.

First of all: I don't think I really understand the exercise at all.

If this is a pushout of topological spaces

then that means $D$ is glued from $B$ and $C$ along $A$; that is, $D \simeq (C \coprod D) /(i(a) \sim j(a) \forall a \in A)$. So the pushout is determined by the maps $i$ and $p$ already. In this case, we have the pushout: $(S^1 \cup \{ \ast \}) / \sim $ where $s \sim \ast $ for all $s \in S^1$. So we identify all points of the boundary with $\ast$.

So constructing $f$ and $j$ is solely for the purpose of making this diagram commutative, isn't it? If so, then for all $s \in S^1$ we need $j(p(s))=j( \ast )=f(i(s))=f(s)$. So $f$ is constant on $S^1$. However, since $f$ needs to be continuous, we can set $f(d)=d$ for all $d \in S^2$. Note that this will send $s$ to $\ast$ for $s \in S^1$ due to $\sim$. Finally we set $j(\ast)=\ast$.

Is this correct?

Second question: Why is $(S^1 \cup \{ \ast \}) / \sim$ equal or homeomorphic to $S^2$ as it says in the diagram?

Answer 1

"Pushout" is a concept of category theory. A pushout diagram associated to two morphisms $i : A \to B$ and $j :A \to C$ with the same domain is a quadratic diagram as in your question having the well-known universal property. Note that the morphisms $i^* : C \to D$ and $j^* : B \to D$ are essential ingredients of a pushout diagram, the "pushout object" $D$ alone does not contain all relevant information. Nevertheless one often reads that $D$ is the pushout of the morphisms $(i,j)$, but that is a very sloppy wording.

The existence of pushouts is a special property of a category. The category $\mathbf{Top}$ of topological spaces and continuous maps has pushouts and they can be constructed as you describe.

Your example is a special case of a pushout of $(i, p)$, where $i : A \to X$ is the inclusion of a subspace and $p : A \to \{*\}$ is the constant map to a one-point space. As the pushout object we get $$D = (X + \{*\})/i(a) \sim p(a) .$$ That is, all $a \in A$ are glued with $*$ so that $$D = X/A$$ is the quotient space of $X$ obtained by collapsing $A$ to a point. The map $i^* : \{*\} \to X/A$ maps $*$ to the common equivalence class of the points of $A$ and $p^* : X \to X/A$ is the quotient map.

Alternatively you can directly check that $\require{AMScd}$ \begin{CD} A @>{i}>> X \\ @V{p}VV @VV{p^*}V \\ \{*\} @>>{i^*}> X/A \end{CD} has the universal property of the pushout.

In your example we get $D^2/S^1$ as the pushout object. It is well-known that $D^2/S^1 \approx S^2$.

So constructing $f$ and $j$ is solely for the purpose of making this diagram commutative, isn't it? If so, then for all $s \in S^1$ we need $j(p(s))=j( \ast )=f(i(s))=f(s)$. So $f$ is constant on $S^1$. However, since $f$ needs to be continuous, we can set $f(d)=d$ for all $d \in S^2$. Note that this will send $s$ to $\ast$ for $s \in S^1$ due to $\sim$. Finally we set $j(\ast)=\ast$.
Is this correct?

Yes until "So $f$ is constant on $S^1$". The next sentence does not make sense. We have $f : D^2 \to S^2$ and thus $f(d)=d$ for all $d \in S^2$ is pointless. .

Second question: Why is $(S^1 \cup \{ \ast \}) / \sim$ equal or homeomorphic to $S^2$ as it says in the diagram?

It is not homeomorphic. You have to take $(D^2 \cup \{ \ast \}) / \sim$.