Does a linear map on a tensor product of vector spaces induce linear maps on the individual spaces? Given vector spaces $U, V$, a linear operator $T$ on $U \otimes V$,
I am aware that $T$ is not necessarily the tensor product of linear operators on $U$ and $V$ respectively.
However, I am wondering if, given a vector $v \in V$, 
do $T$ and $v$ induce a linear operator on $U$?
I suspect this is true (unlike the case for direct sums),
but I can't determine this for sure.
In particular, if $U, V$ have inner products,
$T$ is unitary and $\left\lVert{v}\right\rVert = 1$,
is the induced linear operator unitary?

My Attempt
$U \otimes v := \left\{ u \otimes v \ \middle\vert\  u \in U \right\}$
is a subspace of $U \otimes V$, since it contains $0 \otimes v = 0$,
and $\alpha (u \otimes v) + u' \otimes v = (\alpha u + u') \otimes v$
Define $\pi_1 : U \otimes V \to U$ by $\pi_1(u \otimes v') = u$.
Then 
$$
\pi_1\left(\alpha (u \otimes v') + u' \otimes v'\right) 
= \pi_1\left((\alpha u + u') \otimes v'\right) 
= \alpha u + u' 
= \alpha \pi_1(u \otimes v') + \pi_1(u' \otimes v') ,
$$
so $\pi_1$ is linear.
And define $\tau_v : U \to U \otimes V$
by $\tau_v(u) = u \otimes v$.
Then 
$$
\tau_v(\alpha u + u') 
= (\alpha u + u') \otimes v 
= \alpha (u \otimes v) + (u' \otimes v) ,
$$
so $\tau_v$ is also linear.
Then we can define the induced linear map $T_v$ by
$$
U \xrightarrow{\tau_v} U \otimes v \xrightarrow{T} U \otimes V \xrightarrow{\pi_1} U \ .
$$
I'm a little bit unsure though if this is well-defined.
I'm also not clear whether this may depend on the choice of bases for $U$ and $V$.
Part of the reason I'm not sure about this is because when we replace the tensor product with the direct sum, then the resulting map is affine linear (I believe).

Unitary
Let $U, V$ be inner product spaces and $\lVert \cdot \rVert$ the induced norm.
Then by the standard definition, 
$\left\lVert u \otimes v \right\rVert = \left\lVert u \right\rVert \cdot \left\lVert v \right\rVert$.
If $T$ is a unitary operator on $U \otimes V$ 
and $\left\lvert v \right\rangle \in V$
such that $\left\lVert v \right\rVert = \left\langle v \middle\vert v \right\rangle = 1$,
then $\left\lVert \tau_v(u) \right\rVert = \left\lVert u \right\rVert \left\lVert v \right\rVert = \left\lVert u \right\rVert$
and $T$ does because it is unitary.
But I cannot tell whether $\pi_1$ is.
Is there some way to demonstrate that $\pi_1$ preserves norms?
Alternatively, is there some other way to induce a linear map on $U$?
I suspect it must require the choice of a particular $v \in V$ (but this is okay with me),
otherwise $T$ would factor $S_U \otimes S_V$ into linear operators on $U$ and $V$.
Is there something else I'm missing?
 A: The only thing that doesn't work here is that $\pi_1$ is not well defined. 
Linear maps from $U\otimes V$ to $W$ correspond to bilinear maps from $U\times V$ to $W$.
The correspondence is if $T:U\otimes V\to W$, then $S:U\times V\to W$ is defined by $S(u,v) = T(u\otimes v)$, and conversely, if $S:U\times V\to W$ is bilinear, then $T$ is defined by linear extension from the pure tensors, $T(u\otimes v)=S(u,v)$. 
Now you've defined $\pi_1 : U\otimes V \to U$ on the pure tensors by $\pi_1(u\otimes v)=u$. However, this should correspond to a bilinear map $S:U\times V\to U$ defined by $S(u,v)=\pi_1(u\otimes v)=u$. This is not bilinear, since $S(u,2v)=u=S(u,v)\ne 2S(u,v)$ according to this definition.
However! If we have an inner product, and a chosen $v_0\in V$ (of norm 1), we can fix this! (I assume the second coordinate of inner product is my linear coordinate)
Define 
$S:U\times V\to U$ by
$$S(u,v) = \langle v_0,v\rangle u.$$
You can check that this is bilinear now. Thus it induces a map, I'll call $\pi_{v_0} : U\otimes V \to U$, defined on pure tensors by $\pi_{v_0}(u\otimes v) = \langle v_0,v\rangle u$.
Is $\pi_{v_0}$ unitary?
The answer is no, not in general. No such map can be actually unless $\dim V =1$. The problem is that unitary maps are injective, but if $\dim U = n >0$, $\dim V = m>1$, then 
$\dim U\otimes V = nm > n = \dim U$. Thus no such map can be injective, and thus no such map can be unitary.
However, if $\dim V=1$, then $\pi_{v_0}$ will be unitary.
We check that it preserves the inner product. It suffices to check on the pure tensors. Let $u\otimes v$ and $u'\otimes v'$ be pure tensors. $v=xv_0$, $v'=x'v_0$, for $x$ in the base field, since $V$ is one dimensional.
Then 
$$\langle u\otimes v, u'\otimes v'\rangle 
= \langle u,u'\rangle \langle xv_0,x'v_0\rangle 
$$
$$= \langle u,u'\rangle \overline{x}x'
$$
$$= \langle u,u'\rangle \overline{\langle v_0,xv_0\rangle } \langle v_0,x'v_0\rangle
$$
$$= \langle \langle v_0,v\rangle u, \langle v_0,v'\rangle u'\rangle
$$
$$=\langle \pi_{v_0}(u\otimes v), \pi_{v_0}(u'\otimes v')\rangle,
$$
as desired.
