Let $V$ and $W$ be $k$-vector spaces. Recall that the direct sum of $V$ and $W$ is the vectorspace $V ⊕W =v+w:v\in V,w\in W$. Prove carefully the fact that, $V ⊕W$ can be made into both a product of $V$ and $W$ , and a coproduct of $V$ and $W$ in the category of $k$-vector spaces.
I am hoping someone can help explain what this question is asking to be done.
Cheers
On
I suggest you have a look at the definition of product and coproduct. To make $V \oplus W$ into a product of $V$ and $W$ is to define linear maps $p_1: V \oplus W \to V$ and $p_2: V \oplus W \to W$ (there are obvious candidates) and to verify the universal property: If $X$ is another $k$-vector space and $f_1: X \to V$ and $f_2: X \to W$ are $k$-linear maps, there exists a unique $k$-linear map $f: X \to V \oplus W$ such that $f_i = p_i \circ f$ for $i = 1,2$.
For the coproduct it works similar but with maps $i_1: V \to V \oplus W$ and $i_2: W \to V \oplus W$ into $V \oplus W$. Again there are obvious candidates how to define $i_1, i_2$.
The product of $X$ and $Y$ in the category is the object $X\times Y$ endowed with $p_X:X\times Y\rightarrow X$ and $p_Y:X\times Y\rightarrow Y$ such that for every maps $f:Z\rightarrow X, g:Z\rightarrow Y$ there exists a unique map $h:Z\rightarrow X\times Y$ such that $f=p_X\circ h, g=p_Y\circ h$.
Consider vectors spaces $V,W$, you have a map $p_V:V+W\rightarrow V$ defined by $p_V(v+w)=v$ consider $f:U\rightarrow V, g:U\rightarrow W$, define $(f,g):U\rightarrow V+W$ by $(f,g)(u)=f(u)+g(u)$. $p_V\circ (f,g)(u)=p_V(f(u)+g(u))=f(u)$
The coproduct is the product in the opposite category: the coproduct of $X$ and $Y$ is the object $X+Y$ such that there exist morphisms $i_X:X\rightarrow X+Y$ such that for every norphisms $f:X\rightarrow Z, g:Y\rightarrow Z$ there exists a morphism $h:X+Y\rightarrow Z$ such that $f=h\circ i_X, g=h\circ i_Y$.
Let $i_V:V\rightarrow V+W, i_V(v)=v+0$. Consider maps $f:V\rightarrow U, g:W\rightarrow U$ and define $(f+g):V+W\rightarrow U$ by $(f+g)(v+w)=f(v)+g(w)$.