Rust 当我在 struct 中存储异步函数时，为什么它需要生命周期

#[derive(Default, Serialize, Deserialize, Debug, Eq, Hash, PartialEq)]
pub struct Component {
    name: String,
    module: String,
    r#type: String,
    url: String,
    hash: Option<String>,
    #[serde(skip)]
    retied_times: i8,
}

struct Tasks<'a, F>
where
    F: Future<Output = Result<()>>,
{
    foo: fn(&'a str, &'a Component) -> F,
}

impl<'a, F> Tasks<'a, F>
where
    F: Future<Output = Result<()>>,
{
    pub async fn call(&self, target_dir: &'a str, component: &'a Component) {
        (task)(target_dir, component).await;
    }
}
pub async fn process(target_dir: &str, component: &Component) {
    let task = Tasks { foo: download };
    task.call(target_dir, component);
}

async fn download(target_dir: &str, component: &Component) -> Result<()> {
    //...
}

这段代码将工作得很好.

但是，当我删除 struct Tasks中的生存期'a时，我将不会编译. 编译错误为:

 --> src/lib.rs:28:29
   |
28 |     let task = Tasks { foo: download };
   |                             ^^^^^^^^ one type is more general than the other
   |
   = note: expected fn pointer `for<'r, 's> fn(&'r str, &'s component::Component) -> _`
                 found fn item `for<'r, 's> fn(&'r str, &'s component::Component) -> impl Future<Output = Result<(), anyhow::Error>> {download}`

为什么会发生这种事？