Tensor product commutes with associated graded Let $V,W$ be vector spaces over a field $k$, not necessarily finite-dimensional, and $V_{\bullet}=(V=V_0\supseteq V_1\supseteq\cdots\supseteq V_n=0)$ and $W_{\bullet}=(W=W_0\supseteq W_1\supseteq\cdots\supseteq W_m=0)$ be finite filtrations of each. Then $V\otimes_k W$ admits a natural finite filtration $(V\otimes W)_\bullet$ given by
$$ (V\otimes W)_k=\sum\limits_{i+j=k}V_i\otimes W_j.$$
I want to know whether $\operatorname{gr}(V\otimes W)_\bullet\cong\operatorname{gr}V_\bullet\otimes\operatorname{gr}W_\bullet$ as graded vector spaces, where $\operatorname{gr}$ is the functor which takes a filtered vector space to its associated graded.
More specifically, for each $k$ we have a natural map
$$\phi_k:\bigoplus\limits_{i+j=k}(V_i/V_{i+1})\otimes(W_j/W_{j+1})\to(V\otimes W)_k/(V\otimes W)_{k+1}$$
coming from the natural surjective map
$$\bigoplus\limits_{i+j=k}V_i\otimes W_j\to(V\otimes W)_k/(V\otimes W)_{k+1}$$
coming from the inclusions $V_i\otimes W_j\subseteq(V\otimes W)_k$ when $i+j=k$.  Thus $\phi_k$ is surjective for all $k$.  I would like to know whether $\phi_k$ is always an isomorphism.  I've tried writing out a basis of some splitting of this filtration to show it, and I can't quite complete the argument that way.  
If you have a proof, know a counterexample, or have a reference for this I would really appreciate it!!
 A: One can prove that $\phi_k$ is an isomorphism by noting that it maps a basis of the left hand side to a basis of the right hand side. The following bookkeeping device will be useful to find suitable such bases.

Definition: A filtered basis of the filtered vector space $V_{\bullet}=(V_0\supseteq V_1\supseteq\ldots\supseteq V_n=0)$ is a basis $(e_i)_{i\in I}$ of $V$ whose index set $I$ has a filtration $I=I_0\supseteq I_1\supseteq\ldots\supseteq I_n=\emptyset$ such that $(e_i)_{i\in I_r}$ is a basis of $V_r$, for each $0\leq r\leq n$.

The basis extension theorem ensures that filtered bases always exist.
Now let $V_{\bullet}=(V_0\supseteq V_1\supseteq\ldots\supseteq V_m=0)$ and $W_{\bullet}=(W_0\supseteq W_1\supseteq\ldots\supseteq W_n=0)$ be filtered vector spaces and choose filtered bases $(e_i)_{i\in I}$ and $(f_j)_{j\in J}$ for $V_\bullet$ and $W_{\bullet}$ respectively. The following statements are familiar from linear algebra.

Claim 1:
(a) The family $([e_i])_{i \in I_r\setminus I_{r+1}}$ is a basis of $V_r/V_{r+1}$, where $[e_i]$ denotes the image of $e_i$ under the natural projection $V_r\rightarrow V_r/V_{r+1}$.
(b) The family $([e_i]\otimes [f_j])_{(i,j) \in (I_r\setminus I_{r+1}\times J_s \setminus J_{s+1})}$ is a basis of $V_r/V_{r+1} \otimes W_s/W_{s+1}$.
(c) The family $([e_i]\otimes [f_j])_{(i,j) \in \bigsqcup\limits_{r+s=k}(I_r\setminus I_{r+1}\times J_s\setminus J_{s+1})}$ is a basis of $\bigoplus\limits_{r+s=k}V_r/V_{r+1} \otimes W_s/W_{s+1}$.

We have thus found ourselves a basis of the left hand side. We next determine a basis for the right hand side.

Claim 2:
(a) The family $(e_i\otimes f_j)_{(i,j) \in \bigsqcup\limits_{k\leq r+s\leq m+n}(I_r\setminus I_{r+1}\times J_s\setminus J_{s+1})}$ is a basis of $(V\otimes W)_k$.
(b) The family $([e_i\otimes f_j])_{(i,j) \in \bigsqcup\limits_{r+s=k}(I_r\setminus I_{r+1}\times J_s\setminus J_{s+1})}$ is a basis of $(V\otimes W)_k/(V\otimes W)_{k+1}$.

Proof: (b) follows from (a) using Claim 1.(a), so it suffices to prove (a). Given any $r,s\geq 0$, with $k\leq r+s\leq m+n$, and $(i,j)\in I_r\setminus I_{r+1} \times J_s\setminus J_{s+1}$, we have $e_i\otimes f_j \in (V\otimes W)_k$, by definition. In fact, we have $\bigcup_{r+s=k}I_r\times J_s=\bigsqcup\limits_{k\leq r+s\leq m+n}I_r\setminus I_{r+1}\times J_s\setminus J_{s+1}$, so that the family also spans $(V\otimes W)_k$. Linear independence follows from it being a subfamily of $(e_i\otimes f_j)_{(i,j) \in I\times J}$ which is linearly independent. QED
It remains to observe that the natural morphism
$$
\phi_k: \bigoplus_{r+s=k}(V_r/V_{r+1}\otimes W_s/W_{s+1}) \rightarrow (V\otimes W)_k/(V\otimes W)_{k+1},
$$
mapping $[v]\otimes [w] \mapsto [v\otimes w]$ sends the basis in 1.(c) to the basis in 2.(b) and is therefore an isomorphism.
