# Is this pushout simply connected?

$$\wedge$$ is the wedge product here.

$$S^n$$ is the sphere, $$D^n$$ is the disk of $$n$$-th dimension, so like $$S^{n-1}$$ but filled. Let $$i_1,i_2$$ the inclusions. Let the following diagram be a pushout:

Is $$X$$ simply connected?

I see that all topological spaces beside $$X$$ are simply connected. But I think what we have to do is show that $$X$$ is connected and that $$\pi_1(X)=0$$. For the latter I think we could use van Kampen's theorem. However, I just cannot figure out how do this. First of all, why is $$X$$ connected? And then, how to apply van Kampen? To apply van Kampen, we would need $$((S^2 \times S^1) \wedge S^2) \cup ((S^1 \times D^2)\times S^2) = X$$. But this is not given, right? I just don't see what information at all we have about $$X$$. We have defined push out in a very abstract way as a part of category theory.

If you spoke about pushouts of topological spaces in your lecture, then you will have proven that they exist. This means there is an explicit formula telling you how to construct them. Specifically: the pushout $$X \cup_Y Z$$ of a diagram of topological spaces $$X \overset{f}\leftarrow Y \overset{g}\rightarrow Z$$ is the set $$X \sqcup Z / \sim$$, where $$\sim$$ is the equivalence relation generated by $$f(y)\sim g(y)$$ and the topology is given by the quotient topology.
Let me show that the pushout of nonempty path-connected spaces is again path-connected, to avoid discussing this topology. By construction any two points $$[a],[b] \in X\cup_Y Z$$ are represented by points $$a,b$$ in $$X$$ or $$Z$$. If both live in $$X$$ we can use path-connectedness of $$X$$ to obtain an $$a$$-$$b$$-path in $$X$$, which becomes an $$[a]$$-$$[b]$$-path in the pushout. Similarly if both live in $$Z$$. If $$a\in X$$ and $$b\in Z$$, we can fix an element $$y\in Y$$ and consider a path $$\alpha:a \rightarrow f(y)$$ in $$X$$ and a path $$\beta:g(y)\rightarrow b$$ in $$Z$$. We obtain corresponding paths $$[\alpha]:[a]\rightarrow[f(y)]=[g(y)]$$ and $$[\beta]:[g(y)]\rightarrow[b]$$ in the pushout and I claim that they glue together to one path $$\gamma:[a]\rightarrow[b]$$. One might do this by hand, but as I said, I would like to circumvent describing the topology explicitly. Here the universal property of pushouts comes in handy, since it provides us with continuous maps. Just note that we have a commutative cube and the dashed arrow, provided by the pushout in the back, is a continuous path, which does what we want. (To be completely honest: of cause you have to know that the square in the back is a pushout, which is easy to check by hand)
As a category theorist I would say that being the pushout along inclusions is the very definition of being the union of the objects. But I can see how this might be unintuitive, if one is used to taking unions within some ambient space. In your specific exercise all three spaces canonically embed into $$\Bbb S^2 \times \Bbb D^2 \times \Bbb{S}^2$$ and you can check by hand, that the induced morphism $$X \rightarrow \Bbb S^2 \times \Bbb D^2 \times \Bbb{S}^2$$ is bijective onto the union into this ambient space. By the compact-Hausdorff-trick, this is in fact a homeomorphism. So you might apply SvK in the form you are used to.
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Saying that $$\begin{CD} A @>{i}>> B \\ @V{j}VV @VVV \\ C @>>> D \end{CD}$$ is a pushout of topological spaces is a fancy way to say that $$D$$ is glued from $$B$$ and $$C$$ along $$A$$; that is, $$D \simeq (C \coprod D) /(i(a) \sim j(a) \forall a \in A)$$. It is quite straightforward to check that this satisfies the universal property that is the definition of a pushout.
With this in mind, it should be clear why $$X$$ is connected. It should be also clear how to apply Van Kampen (which with this fancy terminology says that $$\pi_1$$ takes pushout of spaces to pushouts of groups).