Question : For abelian groups $A$ and $B$, define $Hom(A, B)$ to be the set of group homomorphisms $A \rightarrow B$. You may assume that $Hom(A, B)$ itself forms an abelian group under pointwise addition, i. e. , the zero element is the constant zero map $\phi(a) = 0_B$ for all $ a \in A$, for $\phi , \varphi \in Hom(A, B)$ the sum is defined to be the homomorphism given by $(\phi + \varphi)(a) = \phi(a) + \varphi(a) $ for all $a \in A$, and the negative $-\phi$ is the homophism given by$(-\phi)(a) = -\phi(a)$ for all $a \in A$. Prove that: $$ Hom(A \oplus B , C) \cong Hom(A, C) \oplus Hom(B, C)$$ for any abelian groups $A, B, C.$
My approach:
Clearly, $$Hom(A, C), Hom(B, C) \leq Hom(A \oplus B , C) $$ hence $$Hom(A, C) + Hom(B, C) \subseteq Hom(A \oplus B , C) $$ but $$Hom(A, C) + Hom(B, C) \supseteq Hom(A \oplus B , C) $$ since $\forall \phi \in Hom(A \oplus B , C)$ : $\phi(a + b) \rightarrow c \in C$
we have $\phi(a + b) = \phi(a) + \phi(b) \in Hom(A, C) + Hom(B, C)$
Therefore $$Hom(A, C) + Hom(B, C) = Hom(A \oplus B , C) $$
Now we need to show this sum is direct, and it suffices to show $\forall \phi \in Hom(A, C) and \forall \varphi \in Hom(B, C) $, $\phi + \varphi $= zero map from $A \oplus B$ to $C$ will imply both of $\phi$ and $\varphi$ are zero maps.
Suppose we have $$\phi(a) + \varphi(b) = \psi for \forall a \in A \forall b \in B$$,
where $\psi$ maps everything in $A \oplus B$ to zero in $C$. Let $b = 0$ and $a$ ranges over A, we can get $\phi$ is a constant map $\phi(a) = -\varphi(0)$ for all $a$ in $A$. Similarly, $\varphi(b) = \phi(0)$ for all $b$ in $B$.
But since both $\phi$ and $\varphi$ are homomorphisms, we have $\phi(0) = \varphi(0) = 0$.
That implies the sum is direct.
I have just learnt abstract algebra. Did I understand this question correctly at all? Did I make any mistakes? I am not one hundred percent sure. A big thank you to anyone who answers my question.
You are asked to prove $$ \text{Hom}(A \oplus B , C) \cong \text{Hom}(A, C) \oplus \text{Hom}(B, C). $$ When you see a question like this it is almost always sensible to try (i) to find a homomorphism from the LHS to the RHS; (ii) prove that it is one-to-one; (iii) prove that it is onto.
So let's try to find a map $$ F:\text{Hom}(A \oplus B , C) \to \text{Hom}(A, C) \oplus \text{Hom}(B, C). $$ So take an element $\phi\in \text{Hom}(A \oplus B , C)$. We want $F(\phi)\in\text{Hom}(A, C) \oplus \text{Hom}(B, C)$. Recall that $\text{Hom}(A, C) \oplus \text{Hom}(B, C)$ is the set $\{(\theta_1,\theta_2)\mid \theta_1 \in \text{Hom}(A, C), \theta_2 \in \text{Hom}(B, C)\}$.
[Now here is where I use your idea, but write it carefully.] Let's put $F(\phi)=(\theta_1,\theta_2)$, where $\theta_1: a\mapsto \phi(a,0)$, and $\theta_2: b\mapsto \phi(0,b)$. Now $\theta_1:A\to C$ and $\theta_2:B\to C$: but are they in $\text{Hom}(A,C)$ and $\text{Hom}(B,C)$? That is, are they homomorphisms? We must prove that $\theta_1(a_1+a_2)=\theta_1(a_1)+\theta_1(a_2)$, and $\theta_2(b_1+b_2)=\theta_2(b_1)+\theta_2(b_2)$. We must now check that carefully, and I omit the calculations.
So now we have a map $$ F:\text{Hom}(A \oplus B , C) \to \text{Hom}(A, C) \oplus \text{Hom}(B, C); $$ but is it a homomorphism? That is, with the definition we are given of $\phi+\psi$, is $F(\phi+\psi)=F(\phi)+F(\psi)$? This is the case, and it must be carefully checked. Again I omit the details.
Now we must check that $F$ is one-to-one; to do this calculate the kernel. How can $F(\phi)=0$, that is, be the zero-map? By definition this happens only if $\theta_1$ and $\theta_2$ are the zero maps, and now you must check carefully that this means $\phi$ is the zero map. Again I omit the details; you need to supply them.
Finally we must check that $F$ is onto; that is, given any $\psi_1 \in \text{Hom}(A, C)$ and any $\psi_2\in \text{Hom}(B, C)$, can we find a $\phi\in\text{Hom}(A\oplus B, C)$ such that $F(\phi)=(\psi_1,\psi_2)$?
Well the obvious $\phi:A\oplus B\to C$ is the map $(a,b)\mapsto \psi_1(a)+\psi_2(b)$. But is this in $\text{Hom}(A\oplus B\to C)$? You must check carefully that $$ \phi((a_1,b_1)+(a_2,b_2))=\phi(a_1,b_1)+\phi(a_2,b_2). $$ Again I omit the calculations.
When you have supplied all the details you will have a proof. I suggest that when you do so, you justify each step by saying what guarantees the truth of the equality, using such phrases as "definition $+$ on $A\oplus B$", "because $\pi$ is a homomorphism", and so on. At this level you should avoid "clearly" and "obviously"