Is it true that $P_n$ is tight if and only if $P_n^k:=P_n\circ\pi^{-1}_k$ are tight for all k? Let $P_n$ a sequence of probabilies measure on $\mathbb{R}^d$. Define $P_n^k:=P_n\circ\pi^{-1}_k$, where $\pi_k:\mathbb{R}^d\rightarrow\mathbb{R}$ is the k-th projection.
Is it true that $P_n$ is tight if and only if $P_n^k$ are tight for all k?
The implication "only if" is clear. What about the other implication?

If the sequence $\left(P_n\right)_{n\geqslant 1}$ is tight, then for each positive $\varepsilon$, there exists a compact subset set $C$ of $\mathbb R^d$ such that $\inf_{n\geqslant 1}P_n\left(C\right)\gt 1-\varepsilon$. Since $C$ is bounded, we can assume that $C$ has the form $\left[-R, R\right]^d$ for some positive $R$. Otherwise, $C$ is contained in such a set and $\inf_{n\geqslant 1}P_n\left(\left[-R, R\right]^d\right)\geqslant \inf_{n\geqslant 1}P_n\left(C\right)\gt 1-\varepsilon$. 
Tightness of $\left(P_n^k\right)_{n\geqslant 1}$ follows from the inequality   $$P_n^k\left(\left[-R,R\right]\right)= P_n\left\{x =(x_i)_{i=1}^d  , \pi_k (x)\in [-R,R]\right\}\geqslant P_n\left\{x =(x_i)_{i=1}^d  , x_i\in [-R,R]\mbox{ for each }i\in\left\{1,\dots,d\right\}  \right\}.$$
Assume that each sequence $\left(P_n^k\right)_{n\geqslant 1}$ for $1\leqslant  k\leqslant d$ is tight. Let us fix a positive $\varepsilon$. For each $1\leqslant  k\leqslant d$, there exists $R_k\gt 0$ such that for each $n\geqslant 1$, $P_n^k\left(\left[-R_k,R_k\right]\right)\gt 1-\varepsilon/d$. We define $C :=\prod_{k =1}^d \left[-R_k,R_k\right]$. Then 
 equality $$\left\{x=(x_i)_{i=1}^d, x\notin C\right\}=\bigcup_{k=1}^d   \left\{x=(x_i)_{i=1}^d, x_k\notin \left[-R_k,R_k\right]\right\}$$ 
which gives 
$$P_n\left(C^c\right)\leqslant \sum_{k=1}^d P_n\left\{x=(x_i)_{i=1}^d, x_k\notin \left[-R_k,R_k\right]\right\}= \sum_{k=1}^d P_n^k\left(  \left[-R_k,R_k\right]^c\right)\leqslant d\varepsilon/d=\varepsilon.$$
