The natural map $V^* \times W \to \text{Hom}(V,W) $ Is bilinear.

Question

I want to know how to show that $$V^*\times W \to \text{Hom}(V,W) $$ $$(\varphi,w) \mapsto (V \ni v \mapsto \varphi(v) w \in W) $$ is bilinear.

I am currently leaning things again for my exam that is comming up. I know that you need to show bilinearity with.

$f(x + x',y) = f(x,y) + f(x',y)$

$f(x*\alpha, y) = \alpha f(x,y)$

and the same for y as well. Just my issue is that i do not know how we could show that for the $\varphi$ in the map. Can we do something like this ?

$f(\varphi + \mu, w) = (V \to W, (\varphi + \mu)(v)w) = (V \to W, \varphi(v)w + \mu(v)w)= (V\to W, \varphi(v)w) + (V \to W, \mu(v)w) $

Just wanted to make sure i did this right.

All the help is appreciated.

Answer 1

This is written very cumbersome. Nobody writes it like that. I present a way to write down a solution nicely: The issue is that $f(\varphi,w) \in \operatorname{Hom}(V,W)$ is a map, and showing that $$f(\varphi+ \mu,w) = f(\varphi, w) + f(\mu, w)$$ boils down to showing that two maps are equal! But two maps are equal if they are equal on every element of the domain, so for fixed $v \in V$ it suffices to show that $$f(\varphi+\mu,w) (v) = f(\varphi,w)(v) + f(\mu,w)(v)$$

But this last thing is obvious:

$$f(\varphi+ \mu,w)(v) = (\varphi+\mu)(v)w = \varphi(v)w + \mu(v)w = f(\varphi,w)(v) + f(\mu,w)(v)$$

Similarly, you can check all the other axioms for bilinearity.

Answer 2

Your map sends $(\varphi,w)$ to the map $\varphi(\cdot)w \colon v \in V \to \varphi(v)w \in W$. Then $(\rho+\sigma,w)$ is sent to $(\rho+\sigma)(\cdot)w=\rho(\cdot)w+\sigma(\cdot)w$ and $(\varphi,u+w)$ is sent to $\varphi(\cdot)(u+w) = \varphi(\cdot)u+\varphi(\cdot)w$, whence bilinearity.