Set sum of graphs of linear maps is a graph only if the maps are same

Let $S, T \in \text{Hom}(V, W)$. How do you show that the set sum of the graphs of these two maps is itself a graph of a function $f:V \rightarrow W$ only if $S=T$?

I’ve managed to show that $T \in \text{Hom}(V, W)$ iff its graph is a subspace of the Cartesian product $V \times W$. So the set sum of the graphs of $T$ and $S$ is also a subspace of the Cartesian product. Does this help with the proof?

From page 49:

3.6 Let $S$ and $T$ be nonzero linear maps from $V$ to $W.$ The definition of the map $S + T$ is not the same as the set sum of (the graphs of) $S$ and $T$ as subspaces of $V \times W.$ Show that the set sum of (the graphs of) $S$ and $T$ cannot be a graph unless $S = T.$

I can't find the definition in the book, but I presume the set sum of $S$ and $T$ is $\{s + t : s \in S \text{ and } t \in T\}.$

The map $S,$ when identified with its graph, is the set $\{(v, Sv) : v \in V\}$ ; similarly, $T$ is identified with the subset $\{(v, Tv) : v \in V\}$ of $V \times W.$

So, the set sum of $S$ and $T$ is the set $$ \{(v, Sv) + (v', Tv') : v \in V \text{ and } v' \in V\} = \{(v + v', Sv + Tv') : v \in V \text{ and } v' \in V\}. $$ In order for this to be the graph of a function, the set $$ \{Sv + Tv' : v \in V \text{ and } v' \in V \text{ and } v + v' = v''\} $$ must be a singleton, for every vector $v'' \in V.$ In particular, this must be so when $v'' = 0.$ Therefore, the value of the expression $Sv + T(-v) = Sv - Tv$ is constant for all $v \in V.$ Because its value for $v = 0$ is $0,$ we have $Sv - Tv = 0$ for all $v \in V,$ i.e., $Sv - Tv$ for all $v \in V,$ i.e., $S = T.$