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This theorem proof confuses me a little.

$Hom(M \otimes N, P) \cong Hom(M, Hom(N,P))$

In Atiyah book(page no :$28$) it is written that

Any $A$- homomorphism $\phi :M \to Hom_A(N,P)$ defines a bilinear map , namely $(x,y) \to \phi(x) (y)$

My confusion :why is $(x,y) \to \phi(x) y ?$ why not $(x,y) \to \phi(x, y)?$

My thinking : let $f : M \times N \to P $ be a bilinear map defined by $f(x,y) =xy $ where $x \in M$ and $y \in N$

$Hom_A(N, P)$ denote the set of all $A- $module hommomorphism from $N \to P$

Take $g \in Hom_A(N,P)$ then $g : N \to P$

$\phi :M \to Hom_A(N,P) \implies \phi :M \to (g: N \to P) \implies \phi :M \to N \to P $

Therefore $\phi \circ g(x)=f(x,y)$

jasmine
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2 Answers2

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$\phi(x)$ lives in $\mathsf{Hom}_A(N;P)$. $\phi$'s domain is not a product object, it is simply $M$ - so $\phi(x,y)$ doesn't make sense (the arity of $\phi$ is not $2$).

However, since $\phi(x)$ is an $A$-homomorphism $N\to P$, $\phi(x)(y)$ makes sense as an element of $P$. This overall determines a pairing $M\times N\to P$.

$$\phi:M\to(g:N\to P)\implies\phi:M\to N\to P$$

Is wrong, since you are abusing the codomains (twice). $\phi:M\to(g:N\to P)$ isn't correct, nor is $\phi:M\to N\to P$.

FShrike
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By definition, $ϕ$ maps elements from $M$ to elements from $\operatorname{Hom}(N, P)$, which itself is a space of functions.

So if $ϕ(x) = ψ∈\operatorname{Hom}(N, P)$, then $ψ$, evaluated at $y∈N$ is $ψ(y)∈P$. We can skip writing $ψ$ ans sinply write $ϕ(x)(y)$ instead.

Now, one can define a new function $g：M×N→P$, given by $g(x, y) = ϕ(x)(y)$, and show $g$ is bilinear.

Writing $ϕ(x, y)$ is a type-error, because $ϕ$ only takes a single argument $x∈M$.

Hyperplane
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