Can the commutator of two fundamental groups be a fundamental group? Say I have two fundamental groups of nice spaces. There is a natrual map from the free product to the direct product of these groups. 
$$\varphi:\pi_1(X)*\pi_1(Y)\to \pi_1(X)\times\pi_1(Y)$$
This can geometrically be thought of as $\pi_1(X\lor Y)\to\pi_1(X\times Y)$.
The kernel of this homomorphism is $\ker\varphi=[\pi_1(X),\pi_1(Y)]$.
Is there any geometric way of thinking about this commutator? I am hoping it is some $\pi_1$ of some construction of the spaces $X$ and $Y$.
 A: Let $G=\pi_1(X,x_0)$ and $H=\pi_1(Y,y_0)$.  Since $[\pi_1(X),\pi_1(Y)]$ is a subgroup of $\pi_1(X) * \pi_1(Y)$, it is the fundamental group of a certain cover  $Z$ of $X\lor Y$, with the quotient $\pi_1(X)\times \pi_1(Y)$ acting on $Z$ by Deck transformations.
We can describe $Z$ as follows.  Let $p\colon \bigl(\tilde{X},\tilde{x_0}\bigr)\to (X,x_0)$ and $q\colon\bigl(\tilde{Y},\tilde{y_0}\bigr)\to(Y,y_0)$ be universal covers for $X$ and $Y$, respectively.  Then $Z$ is the following subspace of $\tilde{X}\times\tilde{Y}$:
$$
\bigl(\tilde{X}\times q^{-1}(y_0)\bigr) \cup \bigl(p^{-1}(x_0) \times \tilde{Y}\bigr).
$$
Note that $\pi_1(X)\times \pi_1(Y)$ acts on $\tilde{X}\times\tilde{Y}$ componentwise by Deck transformations, and $Z$ is invariant under this action.  It is not hard to see that the quotient of $Z$ by this action is homeomorphic to $X\lor Y$, which proves that it is the desired cover.
For example, if $X$ and $Y$ are circles, then $\tilde{X}=\tilde{Y}=\mathbb{R}$ and $p^{-1}(x_0) = q^{-1}(y_0) = \mathbb{Z}$, so
$$
Z \,=\, (\mathbb{R}\times \mathbb{Z}) \cup (\mathbb{Z} \times \mathbb{R}).
$$
That is, $Z$ is the standard "grid" in the plane.  Note that $Z$ covers $S^1\lor S^1$ in an obvious way, with horizontal lines mapping to the first circle and vertical lines mapping to the second circle.
A: This is what comes to my mind, but it doesn't address your question exactly.  If $\pi_2 X = \pi_2 Y = 0$, then you can take $F$ to be the homotopy fiber of the map $X \vee Y \to X \times Y$.  Then you will have a short exact sequence $$0 \to \pi_1 F \to \pi_1 (X \vee Y) \to \pi_1(X \times Y).$$  In general, you will have $$\pi_2(X \times Y) \to \pi_1 F \to \pi_1 (X \vee Y) \to \pi_1(X \times Y).$$  The idea is that "short exact sequences" in spaces are usually taken to "long exact sequences" in algebra.  
