Given a k-algebra A, representations $V, W$ of $A$, and homomorphism $\phi: V \rightarrow W,$ prove that ker$\phi$ is a subrepresentation of $A$ I think my issue here is wrapping my head around what is, exactly, a homomorphism of representations.
A representation is some vectorspace $S$ and a homomorphism $A \rightarrow$ End$(S)$.
So, to be specific, we have that $V$ is some vectorspace $X$ and a homomorphism $x: A \rightarrow$ End$(X)$. Then $W$ is a vectorspace $Y$ and a homomorphism $y: A \rightarrow$ End$(Y)$.
To make things easier... $V = (x, X)$ and $W = (y, Y)$.
Then... a homomorphism $\phi: V \rightarrow W$ is... what, exactly? It's somehow defined on both the morphisms $x,y$ and the vectorspaces $X,Y$, right? But what does that really "look like?" It  kind of feels like a really, really small functor, but I'm having a hard time putting that into words.
And then, like. Okay. The kernel of any morphism is, colloquially speaking, "the set of things which get mapped to $0$." But I don't know what's really being mapped, here, and what $0$ is in this context. The $0$ representation?
 A: You seem a bit confused. We usually abuse notation and call both the representation and the underlying vector space the same thing. In your question's terminology, $V=X$.
More properly, a representation of a $k$-algebra $A$ is a pair $(V, \rho)$ where $V$ is a $k$-vector space and $\rho \colon A \to \mathrm{End}(V)$ is a $k$-linear ring homomorphism. However, we often abbreviate this by saying "$V$ is a representation of $A$", leaving $\rho$ implicit.
This convention makes more sense when you think of representations as actions: given $v \in V$, we can act on $v$ by $a \in A$ to get another element $a \cdot v \in V$, and this operation is appropriately linear and associative. Formally, $a \cdot v = \rho(a)(v)$ connects the two perspectives. In this sense, a representation "is" the vector space $V$, together with the extra data of an action, but we often just let actions come along for the ride without establishing extra notation for the action map, like scalar multiplication for vector spaces themselves.
A homomorphism of representations then must be a map between the vector spaces that's appropriately compatible with the actions: it's a $k$-linear map $\phi \colon V \to W$ such that $\phi(a \cdot v) = a \cdot \phi(v)$. In this guise it's pretty obvious that $\ker \phi$ is a subrepresentation since $\phi(v)=0$ implies $\phi(a \cdot v) = a \cdot \phi(v) = a \cdot 0 = 0$, giving closure.
You could write the action compatibility axiom more precisely as $\phi(\rho_V(a)(v)) = \rho_W(a)(\phi(v))$. Most people find that impractical.
A: Okay, so I read (part of) a book.
My confusion was coming from the fact that technically a representation is a 2-tuple, and I didn't really understand how the homomorphism was defined on those tuples.
But part of this issue can be solved simply by choosing which part of the representation to refer to. I'll choose the vectorspace part.
So then our homomorphism $\phi: V \rightarrow W$ is defined on the vectorspaces. In other words, $\phi: X \rightarrow Y$, such that $\phi(av) = a \phi(v)$ for all $v \in V_1$. Also, it's linear.
So, there. It's kind of like a "really, really small" functor. Because I say it is.
Now the question of the kernel becomes way mundane. It's all $v \in V_1$ such that $\phi(v) = 0$.
To show that this is a subrepresentation of $V,$ we just have to show that ker$\phi \subset V$ and that $\phi(av) = a\phi(v)$ for all $v \in$ ker$\phi$.
Will update this more after I eat.
