Let's break down and solve the given problem step by step.

We are given:

• $U, V, W$ are vector spaces.
• $T: U \to V$ and $S: V \to W$ are linear maps.
• We need to prove that $\ker(T) \subseteq \ker(S \circ T)$, where $S \circ T: U \to W$ is the composition of $S$ and $T$.

Proof:

1. Recall the definition of the kernel:

   • The kernel of a linear map $T: U \to V$, denoted $\ker(T)$, is the set of all vectors $u \in U$ such that $T(u) = 0$.
   • Similarly, $\ker(S \circ T)$ is the set of vectors $u \in U$ such that $(S \circ T)(u) = 0$.

2. Start with $u \in \ker(T)$:

   • By definition of the kernel, if $u \in \ker(T)$, then $T(u) = 0$.

3. Apply the composition $S \circ T$:

   • Now consider $(S \circ T)(u) = S(T(u))$.
   • Since $T(u) = 0$, this becomes $S(0)$.

4. Use the fact that $S$ is linear:

   • Because $S$ is a linear map, we know that $S(0) = 0$.
   • Therefore, $(S \circ T)(u) = 0$.

5. Conclusion:

   • We have shown that if $u \in \ker(T)$, then $u \in \ker(S \circ T)$.
   • This proves that $\ker(T) \subseteq \ker(S \circ T)$, as required.

Thus, the proof is complete.

Would you like further details or have any questions?

Here are five related questions:

1. What is the composition of two linear maps?
2. How can the rank-nullity theorem apply to this situation?
3. Can we generalize this result to non-linear maps?
4. How does the inclusion $\ker(T) \subseteq \ker(S \circ T)$ relate to injectivity of $T$?
5. How would the proof change if $S$ were not linear?

Tip: When proving set inclusions, always begin by assuming an element belongs to one set and show that it must belong to the other.