I'm struggling with statement (5.2) in Goodwillie's Calculus II: Analytic functors. For a map $Y \to X$ of spaces and a nonempty finite set $T$, Goodwillie defines the "fiberwise join" $Y \ast_X T$ to be the homotopy pushout of $Y \leftarrow Y \times T \to X \times T$. There is a natural map $Y \ast_X T \to X$. In (5.2), Goodwillie says that "clearly"
(5.2): If $Y \to X$ is $(k-1)$-connected, then $Y \ast_X T \to X$ is $k$-connected for nonempty $T$.
This is not clear to me.
Question: How does one see that (5.2) holds?
I can see how to reach the weaker conclusion that $Y\ast_X T \to X$ is $(k-1)$-connected as follows. Note that (5.2) is equivalent to saying that
(5.2, restated): If $Y \to X$ is $(k-1)$-connected, then the square $$\require{AMScd} (\ast) \begin{CD} Y \times T @>>> X \times T\\ @VVV @VVV\\ Y @>>> X \end{CD}$$ is $k$-cocartesian.
Thus one may apply Theorem 2.6 from the same paper. In our case, the relevant facts are that:
The map $X \times T \to X$ is $(-1)$-connected.
The map $Y \to X$ is $(k-1)$-connected.
The square $(\ast)$ is cartesian (i.e. $\infty$-connected).
So Theorem 2.6 tells us that the square $(\ast)$ is $n-1+ \min((-1)+(k-1), \infty)$-connected, with $n=2$. By my arithmetic, this says that $(\ast)$ is $(k-1)$-cocartesian, whereas (5.2) is the stronger claim that $(\ast)$ is $k$-cocartesian.
Unless $T = \varnothing$, then $X \times T \to X$ is actually $0$-connected. Indeed, the induced map $\pi_0(X) \times \pi_0(T) \to \pi_0(X)$ is surjective. (Equivalently, its fiber is $T$, which is $(-1)$-connected... unless it is empty). Which matches up with the assumption in the statement of (5.2), "for nonempty $T$", that you didn't use. So if you do the computation then the square is indeed $k$-connected.