Why does using the default template parameter for a function also used in a lambda parameter not work

Take the following peice of c++ code, which compiles fine (gcc 10.1.0) : -

#include <iostream>
#include <string>
#include <functional>

template <class T = std::string>
void foo(T src, std::function<void(T&& t)> completionFn)
{
    completionFn(std::move(src));
}

int main(int argc, char *argv[])
{
    foo<std::string>("hello", [] (auto && t) {
        std::cout << t << std::endl;
    });

    return 0;
}

If I modify the main function to remove the template parameter in the call to "foo", it no longer compiles even though I have a default template parameter, and I cannot work out why.

int main(int argc, char *argv[])
{
    foo<>("hello", [] (auto && t) {
        std::cout << t << std::endl;
    });

    return 0;
}

I am probably missing something obvious.

Here is the compiler output : -

src/scanner_test.cpp: In function ‘int main(int, char**)’:
src/scanner_test.cpp:19:6: error: no matching function for call to ‘foo(const char [6], main(int, char**)::<lambda(auto:11&&)>)’
   19 |     });
      |      ^
src/scanner_test.cpp:10:6: note: candidate: ‘template<class T> void foo(T, std::function<void(T&&)>)’
   10 | void foo(T src, std::function<void(T&& t)> completionFn)
      |      ^~~
src/scanner_test.cpp:10:6: note:   template argument deduction/substitution failed:
src/scanner_test.cpp:19:6: note:   ‘main(int, char**)::<lambda(auto:11&&)>’ is not derived from ‘std::function<void(T&&)>’
   19 |     });

What am I missing? Thanks! Apologies if it's a silly question.